\section{Introduction}
The behavior of the Riemann zeta function on the critical line has been intensively studied, in particular in relation with the Riemann hypothesis. A natural question concerns the order of magnitude of the moments of $\zeta$:
$$\mu_k(T) := \frac{1}{T} \int_0^T |\zeta (1/2 + it)|^{2k} dt = \mathbb{E} [ |\zeta (1/2 + iUT)|^{2k}],$$
where $k$ is a positive real number and $U$ is a uniform random variable in $[0,1]$.
It is believed that the order of magnitude of $\mu_k(T)$ is $(\log T)^{k^2}$ for fixed $k$ and $T$ tending to
infinity. More precisely, it is conjectured that there exists $C_k > 0$ such that
$$\mu_k(T) \sim_{T \rightarrow \infty} C_k (\log T)^{k^2}.$$
An explicit expression of $C_k$ has been predicting by Keating and Snaith \cite{bib:KSn}, using an expected analogy between $\zeta$ and the characteristic polynomials of random unitary matrices. The conjecture has been only proven for $k =1$ by Hardy and Littlewood and for $k = 2$ by Ingham (see Chapter VII of \cite{Tit}).

The weaker conjecture $\mu_k(T) = T^{o(1)}$ for fixed $k > 0$ and $T \rightarrow \infty$ is equivalent to the Lindel\"of hypothesis which states that $|\zeta(1/2 + it) | \leq t^{o(1)}$ when $t$ goes to infinity. The Lindel\"of hypothesis is a still open conjecture which can be deduced from the Riemann hypothesis.

Under the Riemann hypothesis, it is known that $(\log T)^{k^2}$ is the right order of magnitude for
$\mu_k(T)$. In \cite{Sound}, Soundararajan proves that
$$\mu_k(T) = (\log T)^{k^2 + o(1)}$$
for fixed $k > 0$ and $T$ tending to infinity.
In \cite{Harper}, Harper improves this result by showing that
$$\mu_k(T) \ll_k (\log T)^{k^2},$$
for all $k > 0$ and $T$ large enough, the notation
$A \ll_x B$ meaning that there exists $C > 0$ depending only on $x$ such that $|A| \leq C B$.

On the other hand, the lower bound
$$\mu_k(T) \gg_k (\log T)^{k^2}$$
has been proven by Ramachandra \cite{Ra78, Ra95}) and Heath-Brown \cite{HB81}, assuming the Riemann hypothesis, and, for $k \geq 1$,  by Raziwi\l\l \, and Soundararajan \cite{RS13}, unconditionally.

The moment $\mu_k(T)$ can be written as follows:

$$\mu_k(T) = \mathbb{E} [ \exp ( 2k \Re \log \zeta(1/2 + iUT))].$$
Here $\log \zeta$ denotes the unique determination of the logarithm which is well-defined and countinous everywhere
except at the left of the zeros and the pole of $\zeta$, and which is real on the interval $(1, \infty)$.

It is now natural to also look at similar moments written in terms of the imaginary part of $\log \zeta$:
$$\nu_k (T) =  \mathbb{E} [ \exp ( 2k \Im \log \zeta(1/2 + iUT))].$$
Note that $\Im \log \zeta$ is directly related to the fluctuations of the distribution of the zeros of $\zeta$ with respect to their "expected distribution": we have
$$N(t) = \frac{t}{2 \pi} \log  \frac{t}{2 \pi e} + \frac{1}{\pi} \Im \log \zeta(1/2 + it) + \mathcal{O}(1),$$
where $N(t)$ is the number of zeros of $\zeta$ with imaginary part  between $0$ and $t$.

In the present article, we prove, conditionally on the Riemann hypothesis, an upper bound on $\nu_k(T)$ with the same accuracy as the upper bound on $\mu_k(T)$ obtained by Soundararajan in \cite{Sound}.
The general strategy is similar, by integrating estimates on the tail of the distribution of $\Im \log \zeta$, obtained
by using bounds on moments of sums on primes coming from the logarithm of the Euler product of $\zeta$. The main difference with the paper by Soundararajan \cite{Sound} is that we do  not have an upper bound of $\Im \log \zeta$ which is similar to the upper bound of $\log |\zeta|$ given in his Proposition. On the other hand, from the link between $\Im \log \zeta$ and the distribution of the zeros of $\zeta$, we can deduce that $\Im \log \zeta(1/2 + it)$ cannot decrease too fast when $t$ increases. We intensively use this fact in order to estimate $\Im \log \zeta$ in terms of sums on primes.

The precise statement of our main result is the following:
\begin{theorem}
Under the Riemann hypothesis, for all $k \in \mathbb{R}$, $\varepsilon> 0$,
$$\mathbb{E} [\exp(2 k \Im \log \zeta(1/2 + iTU))] \ll_{k, \varepsilon} (\log T)^{k^2 + \varepsilon},$$
where $U$ is a uniform variable on $[0,1]$.
\end{theorem}
The proof of this result is divided into two main parts. In the first part, we bound the tail of the distribution of $\Im \log \zeta(1/2 + iTU)$ in terms of the tail of an averaged version of this random variable. In the second part, we
show that this averaged version is close to a sum on primes, whose tail is estimated from bounds on its moments. Combining this estimate with the results of the first part gives a proof of the main theorem.

\section{Comparison of $\Im \log \zeta$ with an averaged version}

The imaginary part of $\log \zeta$ varies in a smooth and well-controlled way on the critical line when there are no zeros, and has positive jumps of $\pi$ when there is a zero. We deduce that it cannot decrease too fast.
More precisely, the following holds:

\begin{proposition}
For $2 \leq t_1 \leq t_2$, we have
$$\Im \log \zeta(1/2 + it_2) \geq \Im \log \zeta(1/2 + i t_1) - (t_2 - t_1) \log t_2 + \mathcal{O}(1). $$
\end{proposition}
\begin{proof}
Is is an easy consequence of Theorem 9.3 of Titchmarsh \cite{Tit}: for example, see Proposition 4.1 of \cite{bib:Naj} for details.
\end{proof}

We will now define some averaging of $\Im \log \zeta$ around points of the critical line. From the previous proposition, if
$\Im \log \zeta$ is large at some point $1/2 + it_0$ of the critical line, then it remains large on some segment $[1/2 + it_0,1/2 + i(t_0 + \delta)]$ which tends to also give a large value of an average of $\Im \log \zeta(1/2 + it)$ for $t$ around $t_0$. Our precise way of averaging is the following.
We fix a function $\varphi$ satisfying the following properties: $\varphi$ is real, nonnegative, even, dominated by any negative power at infinity, and its Fourier transform is compactlly supported, takes values in $[0,1]$, is even and equal to $1$ at zero. The Fourier transform is normalized as follows:
$$\widehat{\varphi}(\lambda) = \int_{-\infty}^{\infty} \varphi(x) e^{-i \lambda x} dx.$$
For $H> 0$ we define an averaged version of $\Im \log \zeta$  as follows:
$$I(\tau,H) :=  \int_{-\infty}^{\infty}
\Im \log \zeta (1/2 +  i (\tau + t H^{-1}))\varphi (t) dt.$$
The following result holds:
\begin{proposition}
Let $\varepsilon \in (0,1/2)$. Then, there exist $a, K > 1$, depending only on $\varphi$ and $\varepsilon$, and satisfying the following property.
For $T > 100$, $\tau \in [\sqrt{T},T]$, $ K < V < \log T$, $H := K V^{-1} \log T $, the inequalities
$$\Im \log \zeta(1/2 + i \tau)  \geq V,$$
$$\Im \log \zeta(1/2 + i (\tau - e^r H^{-1}))  \geq - 2  V$$
for all integers $r$ between $0$ and $\log \log T$,
together imply
$$I (\tau + a H^{-1}, H) \geq (1-\varepsilon) V.$$
Similarly, the inequalities
$$\Im \log \zeta(1/2 + i \tau)  \leq -V,$$
$$\Im \log \zeta(1/2 + i (\tau + e^r H^{-1}))  \leq 2 V$$
for all integers $r$ between $0$ and $\log \log T$,
together imply
$$I (\tau - a H^{-1}, H) \leq -(1-\varepsilon) V.$$
\end{proposition}
\begin{proof}
First, we observe that $H > 1$ since $K > 1$ and $V < \log T$. We deduce that for all the values of $s$ such that
$\Im \log \zeta (1/2 + is)$ is explicitly written in the proposition, $ \sqrt{T} - \log T \leq s \leq T + \log T$: in particular
$s > 2$ since $T > 100$, and we can apply the previous proposition to compare these values of $\Im \log \zeta$.

If $\Im \log \zeta (1/2 + i \tau) \geq V$, then
for all $t \geq 0$,
\begin{align*}
& \Im \log \zeta (1/2 + i (\tau + t H^{-1}) ) \geq V - t H^{-1} \log (\tau + tH^{-1})
+ \mathcal{O}(1) \\ &
\geq V - t K^{-1} V (\log T)^{-1} \log (\tau +  t H^{-1}) + \mathcal{O}(1).
\end{align*}
Since $H > 1$ and $\tau \leq T$,
$$\Im \log \zeta (1/2 + i (\tau + t H^{-1}) )
\geq V ( 1 - t K^{-1} (\log T)^{-1} \log(T + t)) + \mathcal{O}(1).$$
We have
$$\log (T + t) = \log T + \log ( 1 + t/T) \leq \log T + \log (1+t),$$
and then, integrating against $\varphi(t-a)$ from $0$ to $\infty$,
\begin{align*}
& \int_0^{\infty} \Im \log \zeta (1/2 + i (\tau + t H^{-1}) ) \varphi(t-a) dt
\\ & \geq V  \left[ \int_{0}^{\infty} \varphi(t-a) dt  -   K^{-1}
\int_{0}^{\infty} t \varphi(t-a) dt \right] +
\mathcal{O}_{a,\varphi}(1),
\end{align*}
since $$V   K^{-1} (\log T)^{-1} \int_0^{\infty} t\varphi(t-a)  \log (1 + t) dt = \mathcal{O}_{a, \varphi}(1),$$
because $V   K^{-1} (\log T)^{-1} < 1$ and $\varphi$ is integrable against $t \log (1+t)$ (it is rapidly decaying at infinity).
We deduce, since $ V K^{-1} > 1$ and then $\mathcal{O}_{a, \varphi}(1) = \mathcal{O}_{a,  \varphi}( V K^{-1})$,
and since the integral of $\varphi$ on $\mathbb{R}$ is $\widehat{\varphi}(0) = 1$,
$$\int_0^{\infty} \Im \log \zeta (1/2 + i (\tau + t H^{-1}) ) \varphi(t-a) dt
\geq V ( 1 - o_{\varphi, a \rightarrow \infty}(1)  -  \mathcal{O}_{a, \varphi}( K^{-1})).$$
If we take $a$ large enough depending on $\varphi$ and $\varepsilon$, and then  $K$ large enough depending on $\varphi$,  $a$ and $\varepsilon$, we
deduce
$$\int_0^{\infty} \Im \log \zeta (1/2 + i (\tau + t H^{-1}) ) \varphi(t-a) dt
\geq V (1 - \varepsilon/2).$$
Now, let us consider the same integral between $-\infty$ and $0$.
For $0 \leq r \leq \log \log T$ integer and $u \in [0, e^{r} - e^{r-1}]$ for $r \geq 1$, $u \in [0,1]$ for $r = 0$,
$$\Im \log \zeta (1/2 + i (\tau - (e^r - u) H^{-1}))
- \Im \log \zeta (1/2 + i (\tau - e^r  H^{-1})) \geq  - u H^{-1} \log T,$$
and then
$$\Im \log \zeta (1/2 + i (\tau - (e^r - u) H^{-1}))
\geq - 2 V - u K^{-1} V$$
Since $u \leq e^r$, $K > 1$, and $ 1 + e^r - u \geq e^{r-1}$,
$$\Im \log \zeta (1/2 + i (\tau - (e^r - u) H^{-1}))
\geq - 2  V  - e (1 + e^r - u) V,$$
and then, for all $t \in [-  e^{\lfloor \log \log T \rfloor}, 0]$,
$$\Im \log \zeta (1/2 + i (\tau + t H^{-1}))
\geq - \mathcal{O}( V (1+|t|)).$$
If  $K$  is  large enough depending on $\varphi$, this estimate remains true for $t <  -  e^{\lfloor \log \log T \rfloor}$, since by Titchmarsh \cite{Tit}, Theorem 9.4, and by the fact that $|t| \geq e^{\log \log T -1 } \gg  \log T$,
\begin{align*}
|\Im \log \zeta (1/2 + i (\tau + t H^{-1}))   | &  \ll
\log ( 2 + \tau + |t| H^{-1})
  \ll \log (T + |t|) \\ & = \log T + \log ( 1 + |t|/T)  \leq \log T + |t|/T \ll |t|,
\end{align*}
whereas $V > K > 1$.
Integrating against $\varphi(t-a)$, we get
$$\int_{-\infty}^0  \Im \log \zeta (1/2 + i (\tau + t H^{-1}) ) \varphi(t-a) dt
\geq - \mathcal{O} ( I V ),$$
where $$I = \int_{-\infty}^0 (1 + |t|)  \varphi(t-a) dt
\leq  \int_{-\infty}^{-a} (1 + |s|)  \varphi(s) ds \underset{a \rightarrow \infty}{\longrightarrow} 0.$$
Hence, for $a$ large enough depending on $\varepsilon$ and $\varphi$,
$$\int_{-\infty}^0  \Im \log \zeta (1/2 + i (\tau + t H^{-1}) ) \varphi(t-a) dt
\geq -  \varepsilon V/2.$$
Adding this integral to the same integral on $[0, \infty)$, we deduce the first part of the proposition.
The second part is proven in the same way, up to minor modifications which are left to the reader.
\end{proof}
In the previous proposition, if we take $\tau$ random and uniformly distributed in $[0,T]$, we deduce the following result:
\begin{proposition}
For  $\varepsilon \in (0,1/2)$, $K$ as in the previous proposition (depending on $\varepsilon$ and $\varphi$), $T > 100$, $K < V < \log T$,
$H = KV^{-1} \log T$, $U$ uniformly distributed on $[0,1]$,
\begin{align*}
\mathbb{P}
[ | & \Im \log \zeta (1/2 + i UT)|  \geq V]
 \leq \mathbb{P} [ |I(UT, H)| \geq (1-\varepsilon) V ]
 \\ & + ( 1 + \log \log T) \mathbb{P}
[ |\Im \log \zeta (1/2 + i UT)| \geq 2 V] + \mathcal{O}_{\varepsilon,\varphi} (T^{-1/2}).
\end{align*}
\end{proposition}

\begin{proof}
We have immediately, by taking $\tau = UT$,
\begin{align*}
\mathbb{P}
[ |\Im \log \zeta (1/2 + i UT)| & \geq V]
 \leq \mathbb{P} [ I(UT + a H^{-1}, H) \geq (1-\varepsilon) V ]
\\ & +  \mathbb{P} [ I(UT - a H^{-1}, H) \leq -(1-\varepsilon) V ]
 \\ & + \sum_{r = 0}^{ \lfloor \log \log T \rfloor} \mathbb{P}
[ \Im \log \zeta (1/2 + i (UT + e^r H^{-1}) ) \geq 2 V]
\\ & + \sum_{r = 0}^{ \lfloor \log \log T \rfloor} \mathbb{P}
[ \Im \log \zeta (1/2 + i (UT - e^r H^{-1}) ) \leq -  2 V]
\\ & + \mathcal{O}(T^{-1/2}),
\end{align*}
the last term being used to discard the event $UT \leq \sqrt{T}$.
Now, for $u \in \mathbb{R} $, the symmetric difference between the uniform laws on $[0,T]$ and $[u H^{-1}, T + u H^{-1}]$
is dominated by a measure of total mass $\mathcal{O} ( |u| H^{-1} T^{-1})$. Hence,
in the previous expression, we can replace $UT + u H^{-1}$ by $UT$ in each event, with the cost of an error term
 $\mathcal{O} ( |u| H^{-1} T^{-1}) = \mathcal{O}(|u| T^{-1})$. The values of $|u|$ which are involved are less than
$\max(a, \log T)$, and there are $\mathcal{O}(\log \log T)$ of them.
Hence, we get an error term $\mathcal{O} (T^{-1}(a + \log T) \log \log T)  = \mathcal{O}_{\varepsilon, \varphi} (T^{-1/2})$ since $a$ depends only on $\varepsilon$ and $\varphi$.
\end{proof}
We can now iterate the proposition: applying it for $V, 2V, 4V,...$. After a few manipulations, it gives the following:
\begin{proposition} \label{sumV}
For $\varepsilon \in (0,1/2)$, $K$ as in the previous proposition (depending on $\varepsilon$ and $\varphi$), $T > 100$, $K< V < \log T$,
$H = KV^{-1} \log T$, $U$ uniformly distributed on $[0,1]$,
\begin{align*}
\mathbb{P}
[ | & \Im \log \zeta (1/2 + i UT)|  \geq V]
   \\ &  \leq \sum_{r = 0}^{p-1} (1 +\log \log T)^r  \mathbb{P} [ |I(UT, 2^{-r} H)| \geq (1-\varepsilon) 2^r V ] +   \mathcal{O}_{\varepsilon,\varphi} (T^{-1/3}),
\end{align*}
where $p$ is the first integer such that $2^p V \geq \log T$.
\end{proposition}
\begin{proof}
We iterate the formula until the value of $V$ reaches $\log T$. The number of steps is dominated by $\log \log T - \log K \leq \log \log T$.  Each step gives an error term
of at most $\mathcal{O}_{\varepsilon, \varphi} ( ( 1+ \log \log T)^{\mathcal{O} (\log \log T)}  T^{-1/2})$. Hence, the total error is $  \mathcal{O}_{\varepsilon,\varphi} (T^{-1/3})$.
We deduce
\begin{align*}
\mathbb{P}
[ |\Im & \log \zeta (1/2 + i UT)|  \geq V] \leq \sum_{r = 0}^{p-1} (1 +\log \log T)^r  \mathbb{P} [ |I(UT, 2^{-r} H)| \geq (1-\varepsilon) 2^r V ]
\\ & + (1 +\log \log T)^p \, \mathbb{P}
[ |  \Im \log \zeta (1/2 + i UT)|  \geq  2^p V] + \mathcal{O}_{\varepsilon, \varphi} (T^{-1/3}),
\end{align*}
Under the Riemann hypothesis, Theorem 14.13 of Titchmarsh \cite{Tit} shows that $|  \Im \log \zeta (1/2 + i UT)|
\ll (\log \log T)^{-1} \log T$ under the Riemann hypothesis. Hence the probability that $|\Im \log \zeta|$ is larger than $2^p V \geq \log T$ is equal to zero if $T$ is large enough, which can be assumed (for small $T$, we can absorb everything in the error term).
\end{proof}
\section{Tail distribution of the averaged version of  $\Im \log \zeta$ and proof of the main theorem}
The averaged version $I(\tau, H)$ of $\Im \log \zeta$ can be written in terms of  sums indexed by primes:

\begin{proposition} \label{Naj32}
Let us assume the Riemann hypothesis. There exists $\alpha > 0$, depending only on the function $\varphi$, such that
for all $\tau \in \mathbb{R}$, $0 < H < \alpha \log  (2+|\tau|)$, $$ I (\tau, H)
 = \Im \sum_{p \in \mathcal{P}} p^{-1/2 - i \tau} \widehat{\varphi} (H^{-1} \log p)
 +
\frac{1}{2} \Im \sum_{p \in \mathcal{P}} p^{-1- 2 i \tau} \widehat{\varphi} (2 H^{-1} \log p) +
\mathcal{O}_{\varphi} (1),
$$
$\mathcal{P}$ being the set of primes.
\end{proposition}
\begin{proof}
This result is an immediate consequence of Proposition 3.2 of \cite{bib:Naj}, which is itself deduced from Lemma 5 of
Tsang \cite{Tsang86}.
\end{proof}
We will now estimate the tail distribution of $I(UT, H)$, where $U$ is uniformly distributed on $[0,1]$, by using upper bounds of the moments of the sums on primes involved in the previous proposition.
 We use Lemma
3 of  Soundararajan \cite{Sound}, which is  presented as a standard mean value estimate by the author (a similar result can be found in Lemma 3.3 of Tsang's thesis \cite{Tsang84}), and which can be stated as follows:
\begin{proposition} \label{Lemma3Sound}
For $T$ large enough and $2 \leq x \leq T$, for $k$ a natural number such that
$x^k \leq T/\log T$, and for any complex numbers $a(p)$ indexed by the primes,
we have
$$\int_{T}^{2T} \left| \sum_{p \leq x, p \in \mathcal{P}} \frac{a(p)}{p^{1/2 + it}} \right|^{2k} dt
\ll T k! \left( \sum_{p \leq x, p \in \mathcal{P}} \frac{|a(p)|^2}{p} \right)^k.$$
\end{proposition}
From Propositions \ref{Naj32} and \ref{Lemma3Sound}, we can deduce the following tail estimate:

\begin{proposition}
For $T$ large enough,  $\varepsilon \in (0,1/10)$, $K > 1$ depending only on $\varepsilon$ and $\varphi$, $0 < V < \log T$,
$H = K V^{-1} \log T$, we have
$$\mathbb{P} [ |I(UT,  H)| \geq (1-\varepsilon) V] \ll_{\varepsilon, \varphi} e^{- (1- 3 \varepsilon) V^2 / \log \log T}
 + e^{-  b_{\varepsilon, \varphi} V \log V} + T^{-1/2},$$
where $b_{\varepsilon, \varphi} > 0$ depends only on $\varepsilon$ and $\varphi$.
\end{proposition}
\begin{proof}
We can assume $V \geq 10 \sqrt{ \log \log T}$ and $V$, $T$ larger than any given quantity depending only on $\varepsilon$ and $\varphi$: otherwise the upper bound is trivial.
In particular, if we choose $\alpha$ as in Proposition \ref{Naj32}, it depends only on $\varphi$, and then, for $K > 1$ depending only on $\varepsilon$ and $\varphi$, we can assume $V > 2K/\alpha$.
From this inequality, we deduce $H  < ( \alpha/2) \log T$, which gives
$H < \alpha \log (2 + UT)$ with probability $  1 - \mathcal{O} (T^{-1/2})$. Under this condition,
 Proposition \ref{Naj32} applies to $\tau = UT$, and we deduce:
$$I(UT,  H) = \Im S_1 + \Im S_2 + \Im S_3 + \mathcal{O}_{\varphi}(1),$$
where
$$S_1 := \sum_{p \in \mathcal{P}, p \leq T^{1/( V \log \log T)} } p^{-1/2 - i UT} \widehat{\varphi} (H^{-1} \log p),$$
$$S_2 := \sum_{p \in \mathcal{P}, p > T^{1/( V \log \log T)} } p^{-1/2 - i UT} \widehat{\varphi} (H^{-1} \log p),$$
$$S_3 := \frac{1}{2} \sum_{p \in \mathcal{P} } p^{-1 - 2 i UT} \widehat{\varphi} (2 H^{-1} \log p).$$
Since we can assume $V$ large depending on $\varepsilon$ and $\varphi$, we can suppose that
the term $ \mathcal{O}_{\varphi}(1)$ is smaller than $\varepsilon V/20$:
$$|I(UT,  H)| \leq |S_1 |+ |S_2| + |S_3 |+ \varepsilon V/20,$$
with probability $  1 - \mathcal{O} (T^{-1/2})$.
We deduce
$$ \mathbb{P} [ |I(UT,  H)| \geq (1-\varepsilon) V] \leq \mathbb{P} [ |S_1| \geq (1- 1.1 \varepsilon) V] $$ $$+
\mathbb{P} [| S_2 |\geq \varepsilon V/100] + \mathbb{P} [| S_3 | \geq \varepsilon V/100]
+ \mathcal{O}(T^{-1/2}).$$

We estimate the tail of these sums by applying Markov inequality to their moment of order $2k$, $k$ being a suitably chosen integer.  Since $\widehat{\varphi}$ is compactly supported, all the sums have finitely many non-zero terms, and
we can apply, for $T$ large, the lemma to all values of $k$ up to $\gg_{\varphi} \log (T /\log T) / H$,
i.e. $\gg_\varphi V/K$, and then $\gg_{\varepsilon, \varphi} V$.
For the sum $S_1$, we can even go up to $(V \log \log T)/2$.
For $S_2$, we can take $k = \lfloor c_{\varepsilon, \varphi} V \rfloor$, for a suitable $c_{\varepsilon, \varphi} > 0$ depending only on $\varepsilon$ and $\varphi$. The moment of order $2k$ is
$$\ll k! \left(\sum_{T^{1/(V \log \log T)} < p \leq  e^{\mathcal{O}_\varphi (H)}} p^{-1} \right)^k
\leq  k^k ( \log \log \log T  + \mathcal{O}_{\varepsilon, \varphi}(1) )^k$$
Hence,
$$\mathbb{P} [ | S_2 |\geq \varepsilon V/100] \leq (\varepsilon V/100)^{- 2 \lfloor c_{\varepsilon, \varphi} V \rfloor}
( c_{\varepsilon, \varphi} V  ( \log \log \log T  + \mathcal{O}_{\varepsilon, \varphi}(1)  ) )^{ \lfloor c_{\varepsilon, \varphi} V \rfloor }
$$
Since we have assumed $V \geq 10 \sqrt{\log \log T}$,
we have
$$(\varepsilon V/100)^{-2}    c_{\varepsilon, \varphi} V  ( \log \log \log T  + \mathcal{O}_{\varepsilon, \varphi}(1)  )
\leq V^{-0.99}$$
and
$$ \lfloor c_{\varepsilon, \varphi} V \rfloor \geq 0.99  \, c_{\varepsilon, \varphi} V,$$
for $T$ large enough depending on $\varepsilon$ and $\varphi$.
Hence
$$ \mathbb{P} [ | S_2| \geq \varepsilon V/100] \leq V^{-0.98 c_{\varepsilon, \varphi} V},$$
which is acceptable.
An exactly similar proof is available for $S_3$, since we even get a $2k$-th moment bounded by $k! (\mathcal{O}(1))^k$.

For $S_1$, the $2k$-th moment is
$$\ll k! (\log \log T + \mathcal{O}_{\varepsilon, \varphi}(1))^k$$
for $k \leq (V \log \log T)/2$. Hence, the probability that $|S_1| \geq W := (1 - 1.1 \varepsilon)V$ is
$$\ll W^{-2k} k! (\log \log T + \mathcal{O}_{\varepsilon, \varphi}(1))^k
$$
We appoximately optimize this expression in $k$. If $V \leq (\log \log T)^2/2$, we can take
$k = \lfloor W^2/ \log \log T \rfloor$ since this expression is smaller than $V \log \log T/2$.  Notice that since
 $V \geq 10 \sqrt{\log \log T}$ and $\varepsilon < 1/10$, we have $W \geq  8 \sqrt{\log \log T}$ and $k$ is strictly positive.
The probability that  $|S_1| \geq W$ is then
$$ \ll [W^{-2} (k/e) (\log \log T +  \mathcal{O}_{\varepsilon, \varphi}(1))]^k \sqrt{k} $$
The quantity inside the bracket is smaller than $e^{-(1- (\varepsilon/100))}$ for $T$ large enough depending on $\varepsilon$ and $\varphi$. Hence, in this case, the probability is
$$ \leq e^{- (1- (\varepsilon/100)) k} \sqrt{k} \ll_{\varepsilon} e^{- (1- (\varepsilon/50)) k}
 \ll e^{ - (1- (\varepsilon/50)) ( 1- 1.1 \varepsilon)^2 V^2 / \log \log T}.$$
This is acceptable.
If $V > (\log \log T)^2/2$, we take $k = \lfloor V \log \log T/2 \rfloor$.
We again get a probability
$$ \ll  [W^{-2} (k/e) (\log \log T +  \mathcal{O}_{\varepsilon, \varphi}(1))]^k \sqrt{k}.$$
Inside the bracket, the quantity is bounded, for $T$ large enough depending on $\varepsilon$ and $\varphi$,
 by
\begin{align*}
& W^{-2} (V \log \log T/2e) (1.001 \log \log T) \leq W^{-2} V   (\log \log T)^2/5.4  \\ & = V^{-1} (\log \log T)^2 (1-1.1\varepsilon)^{-2}/5.4   \leq 2  (1-1.1\varepsilon)^{-2}/5.4 \leq 1/2
\end{align*}
Hence, we get  a probability
$$\ll 2^{-k} \sqrt{k} \ll e^{-k/2}  \ll e^{- V \log \log T/4} \ll e^{-V \log V/4},$$
the last inequality coming from the fact that $V < \log T$ by assumption. This is again acceptable.
\end{proof}
We then get the following bounds for the tail of $\Im \log \zeta$, which easily imply the main theorem by integrating
against $e^{2 k V}$:
\begin{proposition}
For all $\varepsilon \in (0,1/10)$, $V > 0$,
$$\mathbb{P} [ | \Im \log \zeta(1/2 + iUT)| \geq V]
\ll_{\varepsilon}   e^{(\log \log \log T)^3} e^{- (1-\varepsilon) V^2/ \log \log T} + e^{- c_{\varepsilon} V \log V},$$
where $c_{\varepsilon} > 0$ depends only on $\varepsilon$.
\end{proposition}
\begin{proof}
We fix a function $\varphi$ satisfying the assumptions given at the beginning: this function will be considered as universal, and then we will drop all the dependences on $\varphi$ in this proof.
From Theorem 14.13 of Titchmarsh \cite{Tit}, we can assume $V \ll (\log \log T)^{-1} \log T$ and then $V < \log T$ for $T$ large (otherwise the probability is zero).
We can then  also assume $T$ larger than any given quantity depending only on $\varepsilon$ (if $T$ is small, $V$ is small), and $V \geq 10 \sqrt{\log \log T}$. Under these assumptions, we can suppose $V > K$  if $K > 0$ depends only on $\varepsilon$, which allows to apply Proposition \ref{sumV}.
The error term $\mathcal{O}_{\varepsilon}(T^{-1/3})$ can be absorbed in $\mathcal{O}_{\varepsilon}( e^{-c_{\varepsilon} V \log V})$ since $V \ll (\log \log T)^{-1} \log T$. The sum in $r$ is, by the previous proposition, dominated by
\begin{align*}
\sum_{r = 0}^{p-1} (1 + \log \log T)^{r}    e^{- (1-3 \varepsilon)  (2^r V)^2/ \log \log T}
&  + \sum_{r = 0}^{p-1} (1 + \log \log T)^{r} e^{- b_{\varepsilon} (2^r V) \log (2^r V)}
\\ &  +  T^{-1/2} \sum_{r = 0}^{p-1} (1 + \log \log T)^{r},
\end{align*}
where $b_{\varepsilon} > 0$ depends only on $\varepsilon$.
 We can assume $V$ large, and then the exponent in the last exponential decreases by at least $b_{\varepsilon} V$ when $r$ increases by $1$, and then by more than $\log ( 2( 1+ \log \log T))$ when $T$ is large enough depending on $\varepsilon$, since $V \geq 10 \sqrt{ \log \log T}$. Hence, each term of the sum is less than half the previous one and the sum is dominated by its first term. This is absorbed in $\mathcal{O}_{\varepsilon}( e^{-c_{\varepsilon} V \log V})$.

For the last sum, we observe that $2^{p-1}  V < \log T$ by definition of $p$, and then (since we can assume $V > 1$),
$p \ll \log \log T$, which gives a term
$$\ll T^{-1/2} ( \log \log T) (1 + \log \log T)^{ \mathcal{O} (\log \log T)} \ll T^{-1/3},$$
which can again be absorbed in  $\mathcal{O}_{\varepsilon}( e^{-c_{\varepsilon} V \log V})$ since we can assume
$V  \ll (\log \log T)^{-1} \log T$.

For the first sum, we separate the terms for $r \leq 10  \log \log \log T$, and for $r > 10 \log \log \log T$.
For $T$ large, the sum of the terms for $r$ small is at most
\begin{align*} & \sum_{r = 0}^{\lfloor 10 \log \log \log T \rfloor} ( 1 + \log \log T)^{r}
e^{-(1-3 \varepsilon) V^2/ \log \log T}
\\ & \ll (1 + 10 \log \log \log T) e^{ 10 \log \log \log T \log ( 1+ \log \log T)}
e^{-(1-3 \varepsilon) V^2/ \log \log T}
\\ &   \ll
 e^{(\log \log \log T)^3} e^{- (1- 3 \varepsilon) V^2/ \log \log T}
\end{align*}
This term is acceptable after changing the value of $\varepsilon$.
When $r > 10 \log \log \log T$, we have (for $T$ large, $V \geq 10 \sqrt{\log \log T}$ and $\varepsilon < 1/10$)
$$(1-3\varepsilon) (2^r V)^2 / \log \log  T \geq 2^{2r}
= e^{ 20 (\log 2) \log \log \log T}
\geq (\log \log T)^{13}.$$
The exponent is multiplied by $4$ when $r$ increases by $1$, and then decreased by more than $3 (\log \log T)^{13}$, whereas the prefactor is multiplied by $1 + \log \log T$. Hence, the term $r = \lfloor 10 \log \log \log T \rfloor + 1$
dominates the sum of all the terms $r > 10 \log \log \log T$, and its order of magnitude is acceptable.

\end{proof}
\bibliographystyle{plain}


\section{Introduction}

When two black holes spiral together and coalesce, they emit
gravitational radiation which in general possesses net linear
momentum. This accelerates (i.e., ``kicks'') the coalescence remnant
relative to the initial binary center of mass. Analytical calculations
have determined the accumulated kick speed from large separations
until when the holes plunge towards each other
\citep{Per62,Bek73,Fit83,FD84,RR89,W92,FHH04,BQW05,DG06}, but because
the majority of the kick is produced between plunge and coalescence,
fully general relativistic numerical simulations are necessary to
determine the full recoil speed.

Fortunately, the last two years have seen rapid developments in
numerical relativity.  Kick speeds have been reported for non-spinning
black holes with different mass ratios
\citep{HSL06,Baker06,Gonzalez06} and for binaries with spin axes
parallel or antiparallel to the orbital axes
\citep{Herrmann07,Koppitz07, Baker07}, as well as initial explorations
of more general spin orientations
\citep{Gonzalez07,Campanelli07a,Campanelli07b}.  For mergers with low
spin or spins both aligned with the orbital angular momentum, these
results indicate maximum kick speeds $<200$~km~s$^{-1}$.  Remarkably,
however, it has recently been shown that when the spin axes are
oppositely directed and in the orbital plane, and the spin magnitudes
are high (dimensionless angular momentum ${\hat a}\equiv cJ/GM^2\sim
1$), the net kick speed can perhaps be as large as $\sim
4000$~km~s$^{-1}$ \citep{Gonzalez07,sb07,Campanelli07b}.

The difficulty this poses is that the escape speed from most galaxies
is $<1000$~km~s$^{-1}$ (see Figure~2 of \citealt{Merritt04}), and the
escape speed from the central bulge is even smaller.  Therefore, if
large recoil speeds are typical, one might expect that many galaxies
that have undergone major mergers would be without a black hole.  This
is in clear contradiction to the observation that galaxies with bulges
all appear to have central supermassive black holes (see
\citealt{FF05}).  It therefore seems that there is astrophysical
avoidance of the types of supermassive black hole coalescences that
would lead to kicks beyond galactic escape speeds.  From the numerical
relativity results, this could happen if (1)~the spins are all small,
(2)~the mass ratios of coalescing black holes are all much less than
unity, or (3)~the spins tend to align with each other and the orbital
angular momentum.

The low-spin solution is not favored observationally. X-ray
observations of several active galactic nuclei reveal relativistically
broadened Fe K$\alpha$ fluorescence lines indicative of spins ${\hat
a}>0.9$ \citep{Iwasawa96,Fab02,RN03,BR06}. A similarly broad line is
seen in the stacked spectra of active galactic nuclei in a long
exposure of the Lockman Hole \citep{Streb05}. More generally, the
inferred average radiation efficiency of supermassive black holes
suggests that they tend to rotate rapidly (\citealt{Sol82,YT02}; see
\citealt{marconi04} for a discussion of uncertainties).  This is also
consistent with predictions from hierarchical merger models (e.g.,
\citealt{volonteri05}).

Mass ratios much less than unity may occur in some mergers, and if the
masses are different enough then the kick speed can be small.  For
example, \citet{Baker07}, followed by \cite{Campanelli07b}, suggest
that the spin kick component scales with mass ratio $q\equiv
m_1/m_2\leq 1$ as $q^2({\hat a}_2-q{\hat a}_1)/(1+q)^5$, hence for
${\hat a}_1=-{\hat a_2}=1$ the maximum kick speed is $\propto
q^2/(1+q)^4$.  For $q<0.1$ this scales roughly as $q^2$ and hence
kicks are small.  However, for $q>0.2$ the maximum kick is within a
factor $\sim 3$ of the kick possible for $q=1$.  An unlikely
conspiracy would thus seem to be required for the masses always to be
different by the required factor of several.  Some tens of percent of
galaxies appear to have undergone at least one merger with mass ratio
$>0.25$ within redshift $z<1$ (for recent observational results with
different methods, see \citealt{Bell06b,Lotz06}, and for a recent
simulation see \citealt{Maller06}). The well-established tight
correlations between central black hole mass and galactic properties
such as bulge velocity dispersion (see \citealt{FF05} for a review)
then suggest strongly that coalescence of comparable-mass black holes
should be common.

The most likely solution therefore seems to be that astrophysical
processes tend to align the spins of supermassive black holes with the
orbital axis. This astrophysical alignment is the subject of this {\it
  Letter}. Here we show that gas-rich mergers tend to lead to strong
alignment of the spin axes with the orbital angular momentum and thus
to kick speeds much less than the escape speeds of sizeable galaxies.
In contrast, gas-poor mergers show no net tendency for alignment,
assuming an initially uniform distribution of spin and orbital angular
momentum vectors.  We demonstrate this aspect of gas-poor mergers in
\S~2.  In \S~3 we discuss gas-rich mergers, and show that observations
and simulations of nuclear gas in galactic mergers suggest that the
black holes will be aligned efficiently.  We discuss consequences and
predictions of this alignment in \S~4.

\section{Gas-poor Mergers}

Several recent models and observations have been proposed as evidence
that some galactic mergers occur without a significant influence of
gas.  Possible signatures include the metal richness of giant
ellipticals (e.g., \citealt{NO07}) and slow rotation and the presence
of boxy orbits in the centers of some elliptical galaxies (e.g.,
\citealt{Bell06a,Naab06}).

Consider such a gas-free merger, and assume that we can therefore
treat the gradual inspiral of two spinning black holes as an isolated
system.  As laid out clearly by \cite{Schnittman04}, throughout almost
the entire inspiral there is a strong hierarchy of time scales, such
that $t_{\rm inspiral}\gg t_{\rm precess}\gg t_{\rm
orbit}$. \cite{Schnittman04} therefore derived orbit-averaged
equations for the spin evolution in the presence of adiabatic
dissipation. Such effects can lead to relaxation onto favored
orientations. The question is then whether, with the uniform
distribution of orbital and spin directions that seems expected in
galactic mergers, there is a tendency to align in such a way that the
net kicks are small.

Using equations A8 and A10 from \cite{Schnittman04}, we have evolved
the angles between the two spin vectors, and between the spins and the
orbital angular momentum. We find that for isotropically distributed
initial spins and orbits, the spins and orbits at close separation are
also close to isotropically distributed (see
Figure~\ref{fig:finalangles}). Thus, although (as we confirm)
\cite{Schnittman04} showed that for special orientations the spins
might align (e.g., for an initial $\cos\theta_1\approx 1$, or as we
also discovered, for an initial $\cos\theta_2\approx -1$), the initial
conditions resulting in such alignment are special and subtend only a
small solid angle.

The conclusion is that gas-poor mergers alone cannot align spins
sufficiently to avoid large kicks due to gravitational radiation
recoil.  Indeed, \cite{sb07} find that for mass ratios $q>0.25$, spin
magnitudes ${\hat a}_1={\hat a_2}=0.9$, and isotropic spin directions,
$\sim 8$\% of coalescences result in kick speeds $>1000$~km~s$^{-1}$
and $\sim 30$\% yield speeds $>500$~km~s$^{-1}$.  The high maximum
speeds inferred by \cite{Campanelli07b} are likely to increase these
numbers. We now discuss gas-rich mergers, which can naturally reduce
the kick speeds by aligning black hole spins with their orbital axis.

\section{Gas-rich Mergers \label{S_wet_mergers}}

Consider now a gas rich environment, which is common in many galactic
mergers.  The key new element is that gas accretion can exert torques
that change the direction but not the magnitude of the spin of a black
hole, and that the lever arm for these torques can be tens of
thousands of gravitational radii \citep{bp75,np98,na99}.  In
particular, \citet{np98} and \cite{na99} demonstrate that the black
hole can align with the larger scale accretion disk on a timescale
that is as short as 1\% of the accretion time. An important ingredient
of this scenario is the realization by \citet{pp83} that the warps are
transmitted through the disk on a timescale that is shorter by a
factor of $1/2\alpha^2$ compared with the transport of the orbital
angular momentum in flat disks, where $\alpha\sim 0.01-0.1$ is the
standard viscosity parameter \citep{ss73}. The question that
distinguishes gas-rich from gas-poor mergers is therefore whether the
accreted mass is $\sim 0.01-0.1M_{\rm bh}$ during the sinking of the
black holes towards the center of the merged galaxy, where $M_{\rm
bh}$ is a black hole mass.

Numerical simulations show that galactic mergers trigger large gas
inflows into the central kiloparsec, which in gas rich galaxies can
result in a $\sim 10^9 \>{\rm M_{\odot}}$ central gas remnant with a diameter of
only few$\times$100~pc \citep{bh91,bh96,mh94,msh05,kazantzidis05}.
Such mergers are thought to be the progenitors of ultraluminous
infrared galaxies. \citet{kazantzidis05} find that the strong gas
inflows observed in cooling and star formation simulations always
produce a rotationally supported nuclear disk of size $\sim 1-2$~kpc
with peak rotational velocities in the range of 250$-$300 ${\rm km\,
s^{-1}}$.

The results of numerical simulations are in good agreement with
observations, which also show that the total mass of the gas
accumulated in the central region of merger galaxies can reach
$10^9-10^{10}\>{\rm M_{\odot}}$ and in some cases account for about half of the
enclosed dynamical mass \citep{tacconi99}. Observations imply that the
cold, molecular gas settles into a geometrically thick, rotating
structure with velocity gradients similar to these obtained in
simulations and with densities in the range $10^2-10^5\,{\rm cm^{-3}}$
\citep{ds98}. Both observations and simulations of multiphase
interstellar matter with stellar feedback show a broad range of gas
temperatures, where the largest fraction of gas by mass has a
temperature of about $100\,$K \citep{wn01,wn02}.

We therefore consider an idealized model based on these observations
and simulations. In our model, the two black holes are displaced from
the center embedded within the galactic-scale gas disk. We are mainly
concerned with the phase in which the holes are separated by hundreds
of parsecs, hence the enclosed gas and stellar mass greatly exceeds
the black hole masses and we can assume that the black holes interact
independently with the disk. Based on the results of
\citet{escala04,escala05}, \citet{mayer06}, and \citet*{dotti06} the
time for the black holes to sink from these separations to the center
of the disk due to dynamical friction against gaseous and stellar
background is $\leq5\times10^7$yr, which is comparable to the
starburst timescale, $\sim10^8$yr \citep{larson87}.

The accretion onto the holes is mediated by their nuclear accretion
disks fed from the galactic scale gas disk at the Bondi rate, ${\dot
  M}_{\rm Bondi}$, as long as it does not exceed the Eddington rate,
${\dot M_{\rm Edd}}$ \citep{GT04}. Locally, one can estimate the Bondi
radius $R_{\rm Bondi}=GM_{\rm bh}/v_g^2\approx 40~{\rm pc}\,(M_{\rm
  bh}/10^8\,M_\odot)(v_g/100~{\rm km~s}^{-1})^{-2}$ that would be
appropriate for a total gas speed at infinity, relative to a black
hole, of $v_g$ (we use a relatively large scaling of 100~km~s$^{-1}$
for this quantity to be conservative and to include random motions of
gas clouds as well as the small thermal speed within each cloud).  The
accretion rate onto the holes will then be $\min({\dot M_{\rm
    Bondi}},{\dot M_{\rm Edd}})$, where ${\dot M}_{\rm Bondi}\approx
1~M_\odot\,{\rm yr}^{-1}\; (v_g/100\,{\rm
  km\,s}^{-1})^{-3}(n/100\,{\rm cm}^{-3})(M_{\rm bh}/10^8\>{\rm M_{\odot}})^2$.
We also note that clearing of a gap requires accretion of enough gas
to align the holes with the large-scale gas flow.

At this rate, the holes will acquire 1-10\% of their mass in a time
short compared to the time needed for the holes to spiral in towards
the center or the time for a starburst to deplete the supply of gas.
The gas has significant angular momentum relative to the black holes:
analogous simulations in a planetary formation environment suggest
that the circularization radius is some hundredths of the capture
radius (e.g., \citealt{HB91,HB92}). This corresponds to more than
$10^5$ gravitational radii, hence alignment of the black hole spin
axes is efficient (and not antialignment, since the cumulative angular
momentum of the accretion disk is much greater than the angular
momentum of the black holes; see \citealt{king05,lp06}).

If the black hole spins have not been aligned by the time their Bondi
radii overlap and a hole is produced in the disk, further alignment
seems unlikely \citep[for a different interpretation see][]{liu04}. The
reason is twofold: the shrinking of the binary due to circumbinary
torques is likely to occur within $<{\rm few}\times 10^7$~yr
\citep{escala04,escala05}, and accretion across the gap only occurs at
$\sim 10$\% of the rate it would have for a single black hole
\citep{lsa99,ld06,mm06}, with possibly even smaller rates onto the
holes themselves. This therefore leads to the gas-poor merger
scenario, suggesting that massive ellipticals or ellipticals with slow
rotation or boxy orbits have a several percent chance of having
ejected their merged black holes but that other galaxy types will
retain their holes securely.

\section{Predictions, Discussion and Conclusions}

We propose that when two black hole accrete at least $\sim 1-10$\% of
their masses during a gas-rich galactic merger, their spins will align
with the orbital axis and hence the ultimate gravitational radiation
recoil will be $<200$~km~s$^{-1}$. In this section we discuss several
other observational predictions that follow from this scenario.  The
best diagnostic of black hole spin orientation is obtained by
examining AGN jets.  All viable jet formation mechanisms result
in a jet that is initially launched along the spin axis of the black
hole.  This is the case even if the jet is energized by the accretion
disk rather than the black hole spin since the orientation of the
inner accretion disk will be slaved to the black hole spin axis by the
Bardeen-Petterson effect \citep{bp75}.

At first glance, alignment of black hole spin with the large scale
angular momentum of the gas would seem to run contrary to the
observation that Seyfert galaxies have jets that are randomly oriented
relative to their host galaxy disks \citep{kinney00}.  However,
Seyfert morphology is not consistent with recent major mergers
\citep{veilleux03}, hence randomly-oriented minor mergers or internal
processes (e.g., scattering of a giant molecular cloud into the black
hole loss cone) are likely the cause of the current jet directions in
Seyfert galaxies.

Within our scenario, one will never witness dramatic spin orientation
changes during the final phase of black hole coalescence following a
gas-rich merger. There is a class of radio-loud AGN known as
``X-shaped radio galaxies'', however, that possess morphologies
interpreted precisely as a rapid ($<10^5$\,yr) re-alignment of black
hole spin during a binary black hole coalescence
\citep{ekers78,dt02,wang03,komossa03b,lr07,cheung07}. These sources
have relatively normal ``active'' radio lobes (often displaying jets
and hot-spots) but, in addition, have distinct ``wings'' at a
different position angle.  The spin realignment hypothesis argues that
the wings are old radio lobes associated with jets from the one of the
pre-coalescence black holes in which the spin axis possessed an
entirely different orientation to the post-coalescence remnant black
hole \citep{me02}.  If this hypothesis is confirmed by, for example,
catching one of these systems in the small window of time in which
both sets of radio lobes have active hot spots, it would contradict
our scenario unless it can be demonstrated that all such X-shaped
radio-galaxies originate from gas-poor mergers.

However, the existence of a viable alternative mechanism currently
prevents a compelling case from being made that X-shaped radio sources
are a unique signature of mis-aligned black hole coalescences.  The
collision and subsequent lateral expansion of the radio galaxy
backflows can equally well produce the observed wings
\citep{capetti02}.  Indeed, there is circumstantial evidence
supporting the backflow hypothesis.  \citet{kraft05} present a Chandra
observation of the X-shaped radio galaxy 3C~403 and find that the hot
ISM of the host galaxy is strongly elliptical, with a (projected)
eccentricity of $e\sim 0.6$.  Furthermore, the wings of the ``X'' are
closely aligned with the minor axis of the gas distribution,
supporting a model in which the wings correspond to a colliding
backflow that has ``blown'' out of the ISM along the direction of
least resistance.  Although 3C~403 is the only X-shaped radio galaxy
for which high-resolution X-ray maps of the hot ISM are available,
\citet{capetti02} have noted that a number of X-shaped sources have
wings that are oriented along the minor axis of the {\it optical} host
galaxy.  This suggests that the conclusion of \citet{kraft05} for
3C~403 may be more generally true.

There is one particular system, 0402$+$379, that might provide a
direct view of spin alignment in a binary black hole system.  Very
Long Baseline Array (VLBA) imaging of this radio galaxy by
\citet{maness04} discovered two compact flat spectrum radio cores, and
follow-up VLBA observations presented by \citet{rodriguez06} showed
the cores to be stationary.  A binary supermassive black hole is the
most satisfactory explanation for this source, with the projected
distance between the two black holes being only 7.3\,pc.  Within the
context of our gas-rich merger scenario, these two black holes already
have aligned spins.  Existing VLBA data only show a jet associated
with one of the radio cores.  We predict that, if a jet is eventually
found associated with the other radio core, it will have the same
position angle as the existing jet.

Evidence already exists for alignment of a coalescence remnant with
its galactic scale gas disk. \citet{perlman01} imaged the host
galaxies of three compact symmetric objects and discovered nuclear gas
disks approximately normal to the jet axis. The presence of such a
nuclear gas disk as well as disturbances in the outer isophotes of all
three host galaxies suggests that these galaxies had indeed suffered
major gas-rich mergers within the past $10^8$\,yr.

In conclusion, we propose that in the majority of galactic mergers,
torques from gas accretion align the spins of supermassive black holes
and their orbital axis with large-scale gas disks.  This scenario
helps explain the ubiquity of black holes in galaxies despite the
potentially large kicks from gravitational radiation recoil.  Further
observations, particularly of galaxy mergers that do not involve
significant amounts of gas, will test our predictions and may point to
a class of large galaxies without central black holes.

\acknowledgments

We thank Doug Hamilton and Eve Ostriker for insightful discussions.
TB thanks the UMCP-Astronomy Center for Theory and Computation Prize
Fellowship program for support.  CSR and MCM gratefully acknowledge
support from the National Science Foundation under grants AST0205990
(CSR) and AST0607428 (CSR and MCM).


\section*{Introduction}

Let $C$ be a nonempty convex set in a (real) normed linear space $X$. A
function $f\colon C\to\mathbb{R}$ is called {\em d.c.}\ (or ``delta-convex'')
if it can be represented as a difference of two continuous convex
functions on $C$. An extension of this notion, the notion of a {\em d.c.\ mapping}
$F\colon C\to Y$ (see Definition~\ref{D:dc}) where $Y$ is a normed linear space,
was introduced in \cite{VeZa1} and studied in \cite{VeZa1},
\cite{DuVeZa}, \cite{VeZa2} and some other papers by the authors.

The present paper concerns the following natural questions.
\begin{enumerate}
\item[(Q1)] When is it possible to extend a d.c.\ function (or a d.c.\ mapping) on $C$
to a d.c.\ function (or a d.c.\ mapping) on the whole $X$?
\item[(Q2)] When is it possible to extend a continuous convex function
on a closed subspace $Y$
of $X$ to a continuous convex function on $X$?
\end{enumerate}

In Section 2, we show how results of \cite{VeZa2} on compositions of
d.c.\ functions and mappings imply positive results concerning (Q1).
For instance, Corollary~\ref{elpecka}(a) reads as follows.

{\em
Let $X$ be a (subspace of some) $L_p(\mu)$ space with $1<p\le2$. Let
$C\subset X$ be a convex set with a nonempty interior. Then each continuous
convex function  $f$ on $C$, which is Lipschitz on every bounded subset of
$\mathrm{int}\,C$, admits a d.c.\ extension to the whole $X$.}

\noindent
(Note that only the
case of $C$ unbounded is interesting; cf.~Lemma~\ref{F:1}(c).)
The needed results from \cite{VeZa2}, together with some definitions and
auxiliary facts, are presented in Section~1 (Preliminaries).

Section 3 contains two counterexamples.
The first one (Example~\ref{pasovec})
shows that, in the above mentioned
Corollary~\ref{elpecka}(a), we cannot conclude that $f$ admits a
continuous {\em convex} extension (even for $X= \mathbb{R}^2$).
The second counterexample (Example~\ref{koulovec}) shows that,
in the above mentioned Corollary~\ref{elpecka}(a),
it is not possible
to relax the assumption that $f$ is Lipschitz on bounded sets by assuming that $f$ is only
locally Lipschitz on $C$.

In the last Section~4, we consider the question (Q2)
of extendability of
continuous convex functions from a closed subspace $Y$ to the whole $X$.
The authors of~\cite{BMV} obtained a necessary and
sufficient condition on $Y$ in terms of nets in $Y^*$ and, using
Rosenthal's extension theorem, they proved the following interesting
corollary (\cite[Corollary~4.10]{BMV}).

{\em If $X$ is a Banach space and $X/Y$ is separable, then each continuous convex function on
$Y$ admits a continuous convex extension to $X$.}

\noindent
 Using methods from \cite{H} and \cite{VeZa2}, we give a
 necessary and sufficient condition on $Y$ of a different type in Theorem~\ref{univ}.
As an application, we present an
elementary alternative proof of the above mentioned
  \cite[Corollary~4.10]{BMV}, which works also for noncomplete $X$.

\section{Preliminaries}\label{S:prelim}

We consider only normed linear spaces over the
reals $\mathbb{R}$. For a normed linear space $X$ we use the following
fairly standard notations:
$B_X$ denotes the closed unit ball; $U(c,r)$ is the open ball centered in $c$ with radius $r$;
$[x,y]$ is the closed segment $\mathrm{conv}\{x,y\}$ (the meaning of the symbols
$(x,y)$ and $(x,y]=[y,x)$ is clear). By definition, the distance of a
set from the empty set $\emptyset$ is $\infty$, and the restriction of a
mapping to $\emptyset$ has all properties like continuity, Lipschitz
property, boundedness and so on.

We will frequently use also the following less standard notation.

\begin{notation}
Let $A,B,A_n,B_n$ ($n\in\mathbb{N}$) be subsets of a normed linear space $X$. We
shall write:
\begin{itemize}
\item $A\subset\subset B$ whenever there exists $\epsilon>0$
such that $A+\epsilon B_X\subset B$;
\item $A_n\nearrow A$ whenever $A_n\subset A_{n+1}$ for each
$n\in\mathbb{N}$, and $\bigcup_{n\in\mathbb{N}}A_n=A$;
\item $A_n\nearrow\!\!\!\nearrow A$ whenever $A_n\subset\subset A_{n+1}$ for each
$n\in\mathbb{N}$, and $\bigcup_{n\in\mathbb{N}}A_n=A$.
\end{itemize}
\end{notation}

We shall use the following simple facts about convex sets and functions.

\begin{lemma}[{\cite[Lemma~2.3]{VeZa2}}]\label{L:dn}
Let $C\subset X$ be nonempty, open and convex. Let $\{C_n\}$ be a sequence of
convex sets with nonempty interiors, such that $C_n\nearrow C$.
Then there exists a sequence $\{D_n\}$ of nonempty,
bounded, open,
convex sets such that $D_n\nearrow\!\!\!\nearrow C$, and $D_n\subset\subset C_n$ for each $n$.
\end{lemma}

\begin{lemma}[{\cite[Fact~1.6]{VeZa2}}]\label{F:1}
Let $C\subset X$ be a nonempty convex set, $f\colon C\to\mathbb{R}$ be a
convex function.
\begin{enumerate}
\item[(a)] If $C$ is open and bounded and $f$ is continuous,
           then $f$ is bounded below on $C$.
\item[(b)] If $f$ is bounded on $C$ then $f$ is Lipschitz on each $D\subset\subset C$.
\item[(c)] If $f$ is Lipschitz then it admits a Lipschitz convex
extension to $X$.
\end{enumerate}
\end{lemma}

\begin{lemma}\label{K}
Let $f$ be a continuous convex function on an open convex subset $C$ of
a normed linear space $X$. Then there exists a sequence $\{D_n\}$ of
nonempty bounded open convex sets such that $D_n\nearrow C$ and $f$ is
Lipschitz (and hence bounded) on each $D_n$.
\end{lemma}

\begin{proof}
Fix $x_0\in C$ and consider the nonempty open convex sets
$C_n:=\{x\in C: f(x)<f(x_0)+n\}$. By Lemma~\ref{L:dn}, there exist
nonempty bounded open convex sets $D_n$ such that $D_n\subset\subset C_n$ and
$D_n\nearrow\!\!\!\nearrow C$. Using Lemma~\ref{F:1}(a), it is easy to see that $f$ is
bounded on each $D_{n+1}$. Hence, by Lemma~\ref{F:1}(b), $f$ is Lipschitz
on each $D_n$.
\end{proof}

Let us recall the following easy known fact (see, e.g., \cite[Theorem~1.25]{valentine}):
if $A,B$ are convex sets in
a vector space then
\begin{equation}\label{Val}
\mathrm{conv}(A\cup B)=\bigcup_{0\le t\le1}[(1-t)A+tB]=\bigcup_{a\in A,\, b\in B}[a,b]\,.
\end{equation}

\begin{lemma}\label{conv}
Let $Y$ be a closed subspace of a normed linear space $X$,
$C\subset Y$ and $A\subset X$ convex sets.
\begin{enumerate}
\item[(a)] $\mathrm{conv}(A\cup C)\cap Y=\mathrm{conv}[(Y\cap A)\cup C]$.
\item[(b)] If $\mathrm{int}\,A\ne\emptyset$ and $A$ is dense in $X$,
then $A=X$.
\item[(c)] If $C$ is open in $Y$, $A$ is open in $X$ and
$A\cap C\ne\emptyset$, then $\mathrm{conv}(A\cup C)$ is open.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The inclusion ``$\supset$'' is obvious. To prove the other
inclusion, consider an arbitrary $y\in Y\cap\mathrm{conv}(A\cup C)$.
Then $y\in[a,c]$ for some $a\in A$, $c\in C$. If $y\ne c$ then
necessarily $a\in Y$ (since $y,c\in Y$) and hence
$y\in \mathrm{conv}[(Y\cap A)\cup C]$; and the last formula is trivial
for $y=c$.\\
(b) follows, e.g., from the well-known fact that
$\mathrm{int}(\overline{A})=\mathrm{int}\,A$ whenever
$\mathrm{int}\,A$ is nonempty.\\
(c) Fix an arbitrary $a_0\in A\cap C$. For each $x\in C$, there
obviously exists $y\in C\setminus\{a_0\}$ such that $x\in(y,a_0]$;
consequently, there exists
$t\in(0,1]$ with $x\in(1-t)C+tA$. Now we are done, since
$$\mathrm{conv}(A\cup C)=
C\cup\bigcup_{0<t\le1}[(1-t)C+tA]=
\bigcup_{0<t\le1}[(1-t)C+tA]$$
and the members of the last union are open.
\end{proof}

In the rest of this section, we collect some facts about d.c.\ functions
and mappings, which we will need in the next sections.

Let $C$ be a convex set in a normed linear space $X$. Recall that a
function $f\colon C\to\mathbb{R}$ is {\em d.c.}\ (or ``delta-convex'')
if it can be represented as a difference of two continuous convex
functions on $C$. The following generalization to the case of vector-valued
mappings on $C$ was studied in \cite{VeZa1} for open $C$, and in
\cite{VeZa2} for a general (convex) $C$.

\begin{definition}\label{D:dc}
Let $X,Y$ be normed linear spaces, $C\subset X$ be a convex set, and
$F\colon C\to Y$ be a continuous mapping. We say that $F$ is {\em d.c.}\
(or ``delta-convex'') if there exists a continuous (necessarily convex)
function $f\colon C\to\mathbb{R}$ such that $y^*\circ F+f$ is convex on $C$
whenever $y^*\in Y^*$, $\|y^*\|\le1$. In this case we say that $f$
controls $F$, or that $f$ is a {\em control function} for $F$.
\end{definition}

\begin{remark}\label{R:dc}
It is easy to see (cf.~\cite{VeZa1}) that:
\begin{enumerate}
\item[(a)] a mapping
$F=(F_1,\ldots,F_m)\colon C\to\mathbb{R}^m$ is d.c.\ if and only if each of its
components $F_k$ is a d.c.\ function;
\item[(b)] the notion of delta-convexity does not depend on the choice
of equivalent norms on $X$ and $Y$.
\end{enumerate}
\end{remark}

\begin{lemma}[{\cite[Lemma~5.1]{VeZa2}}]\label{ndc}
Let $X,Y$ be normed linear spaces, let $A\subset X$ be an open convex
set with $0\in A$, and let $F\colon A\to Y$ be a mapping. Suppose there exist
$\lambda\in(0,1)$ and a sequence of balls $B(x_n,\delta_n)\subset A$ such
that $\{x_n\}\subset\lambda A$, $\delta_n\to0$ and $F$ is unbounded on
each $B(x_n,\delta_n)$. Then $F$ is not d.c.\ on $A$.
\end{lemma}

The following result was proved in \cite[Theorem~18(i)]{DuVeZa} for
$X,Y$ Banach spaces, but the proof therein works for normed linear
spaces as well.

\begin{proposition}\label{P:lip}
Let $X,Y$ be normed linear spaces, $C\subset X$ be a bounded open convex
set, and $F\colon C\to Y$ be a d.c.\ mapping with a Lipschitz control
function. Then $F$ is Lipschitz.
\end{proposition}

\begin{lemma}[{\cite[Lemma~2.1]{VeZa2}}]\label{L:hartman}
Let $X,Y$ be normed linear spaces, $C\subset X$ a nonempty
convex set, and  $F\colon C\to Y$ a mapping. Let $\emptyset \neq D_n \subset C$
($n\in\mathbb{N}$)
be convex sets such that $D_n\nearrow C$ and, for each $n$,
$\mathrm{dist}(D_n,C\setminus D_{n+1})>0$,
$D_n$ is relatively open in $C$,
and
$F|_{D_n}$ is d.c.\ with a
control function  $\gamma_n\colon D_n\to \mathbb{R}$
which is either bounded or Lipschitz. Then $F$ is d.c.\ on $C$.
\end{lemma}

An important ingredient of the present paper is application of the
following two results on compositions of d.c.\ mappings.

\begin{proposition}[\cite{VeZa1},\cite{VeZa2}]\label{P:veza}
Let $X,Y,Z$ be normed linear spaces, $A\subset X$ and $B\subset Y$
convex sets, and $F\colon A\to B$ and $G\colon B\to Z$ d.c.\ mappings.
If $G$ is Lipschitz and has a Lipschitz control function,
then $G\circ F$ is d.c.\ on $A$.
\end{proposition}

\begin{lemma}[{\cite[Lemma~3.2(ii)]{VeZa2}}]\label{P:main}
Let $U,V,W$ be normed linear spaces, let $A\subset U$ be an open convex
set
and $B\subset V$ a convex set, and let $\Phi\colon A\to B$
 and $\Psi\colon B\to W$ be mappings.
Suppose that $\Phi$ is d.c.\ and
there exist sequences of convex sets $A_n \subset A$, $B_n \subset B$
such that
$\mathrm{int}\,A_n \neq \emptyset$,
$A_n\nearrow A$,
$\Phi(A_n) \subset B_n$, and $\Psi|_{B_n}$ is Lipschitz and d.c.\ with a
Lipschitz control function.
 Then $\Psi\circ \Phi$ is d.c.\ on $A$.
 \end{lemma}

Let us recall that a normed linear space $X$ is said to have {\em
modulus of convexity of power type 2} if there exists $a>0$ such that $\delta_X(\epsilon)\ge
a\epsilon^2$ for each $\epsilon\in(0,2]$ (where $\delta_X$ denotes the
classical modulus of convexity of $X$; see e.g.\ \cite[p.409]{BeLin} for the
definition).

\begin{proposition}[{\cite[Corollary~3.9(a)]{VeZa2}}]\label{bilinear}
Let $Y,V,X,Z$ be normed linear spa\-ces and let both
$Y$ and $V$ admit renormings with modulus of
convexity of power type 2. Let $B\colon Y\times V\to Z$ be a
continuous bilinear mapping,
 $C\subset X$  an open convex set, and let
$F\colon C\to Y$ and $G\colon C\to V$ be d.c.\ mappings. Then the
mapping $B\circ(F,G)\colon x\mapsto B(F(x),G(x))$ is d.c.\ on $C$.
\end{proposition}

\section{Extensions of d.c.\ mappings}

Let $X,Y$ be normed linear spaces, $C\subset X$ be a convex set, and
$F\colon C\to Y$ be a d.c.\ mapping.
In the present section, we are interested in existence of a d.c.\
extension of $F$ to the whole $X$ or at least to the closure of $C$. Let
us start with a simple observation.

\begin{observation}\label{O:exten}
Let $X,Y,C,F$ be as in the beginning of this section, and
$f\colon C\to\mathbb{R}$ a control function of $F$.
\begin{enumerate}
\item[(a)] If $Y$ is finite-dimensional, and both $F,f$ are Lipschitz on
$C$, then $F$ admits a d.c.\ extension to $X$.
\item[(b)] If both $F,f$ admit continuous extensions $\tilde{F},\tilde{f}$
to a convex set $D$ such that $C\subset D\subset\overline{C}$,
then $\tilde F$ is d.c.\ with the control function $\tilde f$.
\end{enumerate}
\end{observation}

\begin{proof}
 By Remark~\ref{R:dc}, it suffices to prove {\it(a)} for $Y=\mathbb{R}$. In this
case, $F=(F+f)-f$ is a difference of two Lipschitz convex functions on
$C$. By Lemma~\ref{F:1}(c), $F$ can be extended to a difference of two
Lipschitz convex functions on $X$. The assertion
{\it(b)} follows by a simple limit argument.
\end{proof}

\begin{proposition}\label{L:uzaver}
Let $X$ be a normed linear space, $Y$ a Banach space, $C\subset X$
a convex set with a nonempty interior and $F\colon C\to Y$ a d.c.\ mapping.
Suppose there exists a nondecreasing sequence $\{A_n\}$ of open convex
sets in $X$ such that $\overline{C}\subset \bigcup A_n$ and, for each
$n$, $F|_{(\mathrm{int}\,C)\cap A_n}$ has a Lipschitz control function.
Then $F$ admits a
d.c.\ extension to $\overline{C}$.
\end{proposition}

\begin{proof}
Since $A_n\nearrow A:=\bigcup_k A_k$,
Lemma~\ref{L:dn} allows us to suppose that the sets $A_n$ are also
bounded and satisfy $A_n\nearrow\!\!\!\nearrow A$. By Proposition~\ref{P:lip},
$F|_{(\mathrm{int}\,C)\cap A_n}$ is Lipschitz for each $n$. Consequently,
since $\overline{C}\cap A_n\subset\overline{(\mathrm{int}\,C)\cap A_n}\;$,
$F|_{(\mathrm{int}\,C)\cap A_n}$ has
a unique Lipschitz extension $F_n^*\colon \overline{C}\cap A_n\to Y$.
Since, for $n_2>n_1$, $F^*_{n_2}$ obviously extends $F^*_{n_1}$, there
exists a unique continuous $F^*\colon \overline{C}\to Y$ which
extends each $F^*_n$. Since, for each $n$,
any Lipschitz control function for $F|_{(\mathrm{int}\,C)\cap A_n}$
has a Lipschitz extension to $D_n:=\overline{C}\cap A_n$,
Observation~\ref{O:exten}(b) gives that $F^*|_{D_n}$ has a Lipschitz control
function. Moreover,
$\bigcup D_n=\overline{C}$ and
$\mathrm{dist}(D_n,\overline{C}\setminus D_{n+1})=
\mathrm{dist}(D_n,\overline{C}\setminus A_{n+1})\ge
\mathrm{dist}(A_n,X\setminus A_{n+1})>0$ for each $n$. Applying
Lemma~\ref{L:hartman} with $D:=\overline{C}$, we obtain that $F^*$ is
d.c.\ on $\overline{C}$.
\end{proof}

Now we will prove the main result of the present section.
For the definition of modulus of convexity of power type 2
see the text before Proposition \ref{bilinear}.

\begin{theorem}\label{T:rozsir}
Let $X,Y$ be normed linear spaces, $C\subset X$ a convex set with a
nonempty interior and $F\colon C\to Y$  a d.c.\ mapping. Let
$A\supset C$ be an open convex set in $X$.
Suppose that
$X$ admits a renorming with modulus of convexity of
power type 2, and either $C$ is closed or $Y$ is a Banach space.
Then the following assertions are equivalent.
\begin{enumerate}
\item[(i)] $F$ admits a d.c.\ extension $\widehat{F}\colon A\to Y$.
\item[(ii)] Some control function $f$ of $F$ admits a continuous convex extension
$\widehat{f}\colon A\to \mathbb{R}$.
\item[(iii)] There exists a nondecreasing sequence $\{D_n\}$ of open convex sets
such that $A=\bigcup D_n$ and,
for each $n$,
$(\mathrm{int}\,C)\cap D_n\ne\emptyset$ and
the restriction of $F$ to
$(\mathrm{int}\,C)\cap D_n$ has a Lipschitz control function.
\end{enumerate}
\end{theorem}

\begin{proof}
The implication $(i)\Rightarrow(ii)$ is trivial, while
$(ii)\Rightarrow(iii)$ follows immediately from Lemma~\ref{K} applied to $\hat f$.

Let us prove $(iii)\Rightarrow(i)$.
By translation we can suppose that $0\in (\mathrm{int}\,C)\cap D_1$. The
sets $A_n:=D_n\cap U(0,n)$ ($n\in\mathbb{N}$) form a sequnce of
bounded open convex sets such that
$A_n\nearrow A$, $0\in (\mathrm{int}\,C)\cap A_1$ and, for each $n$,
\begin{equation}\label{contr}
\text{
$F|_{(\mathrm{int}\,C)\cap A_n}$ has a Lipschitz control function $f_n$.
}
\end{equation}
First we will extend $F$ to a mapping
$F^*\colon \overline{C}\cap A\to Y$. If $C$ is closed, then
$\overline{C}\cap A=C$; so we put $F^*=F$. If $C$ is not closed,
$Y$ is a Banach space by the assumptions. Proposition~\ref{P:lip} and
\eqref{contr} imply that $F$ is Lipschitz on
$(\mathrm{int}\,C)\cap A_n$.
Note that $\overline{C}\cap A_n\subset\overline{(\mathrm{int}\,C)\cap A_n}\,$;
thus $F|_{(\mathrm{int}\,C)\cap A_n}$ has
a unique Lipschitz extension $F_n^*\colon \overline{C}\cap A_n\to Y$.
Since, for $n_2>n_1$, $F^*_{n_2}$ obviously extends $F^*_{n_1}$, there
exists a unique continuous $F^*\colon \overline{C}\cap A\to Y$ which
extends each $F^*_n$.

In both cases ($C$ closed or not), $f_n$ has a Lipschitz extension to
$B_n:=\overline{C}\cap A_n$. By Observation~\ref{O:exten}(b),
\begin{equation}\label{contr2}
\text{
$F^*|_{B_n}$ is Lipschitz and d.c.\ with a Lipschitz control function.
}
\end{equation}

Denote by $\mu$ the Minkowski functional of $C$,
i.e.
\[
\mu(x)=\inf\{t>0: x\in tC\}.
\]
It is well known that $\mu$ is a Lipschitz convex function on $X$
(recall that $0\in\mathrm{int}\,C$), and
$\mu(x)\le1$ if{f} $x\in \overline{C}$. Consider the ``radial projection'' $P$ onto
$\overline{C}$, given by
\[
P(x)=\begin{cases}
x &\text{if $x\in \overline{C}$;}\\
\frac{x}{\mu(x)} &\text{if $x\in X\setminus \overline{C}$.}
\end{cases}
\]
The function $x\mapsto\max\{1,\mu(x)\}$ is convex and Lipschitz, and its
values belong to $[1,\infty)$. The function $t\mapsto\frac{1}{t}$ is
convex and Lipschitz on $[1,\infty)$; consequently, by
Proposition~\ref{P:veza}, the composed function $x\mapsto\frac{1}{\max\{1,\mu(x)\}}$
is d.c.\ on $X$. Moreover, the mapping $B\colon\mathbb{R}\times X\to X$, given
by
\[
B(t,x)=tx\,,
\]
is a continuous bilinear mapping. Since
$P(x)=B\left(\frac{1}{\max\{1,\mu(x)\}},x\right)$, $x\in X$,
 Proposition~\ref{bilinear} implies
that $P$ is d.c.\ on $X$.

Let us show that $\hat{F}:=F^*\circ(P|_A)$ is a d.c.\ extension of $F$
to $A$. The fact that $\hat{F}$ extends $F$ is obvious. To prove that
$\hat{F}$ is d.c., it is sufficient to apply Lemma~\ref{P:main} with
$B:=\overline{C}\cap A$, $\Phi:=P|_A$, $\Psi:=F^*$ (and $A,A_n,B_n$ as above).
Indeed, the assumptions of that lemma are satisfied since
$\Phi(A_n)=P(A_n)\subset \overline{C}\cap A_n=B_n$
(note that $P(A_n)\subset A_n$ because $0\in A_1$) and \eqref{contr2}
holds.
\end{proof}

\begin{remark}
\begin{enumerate}
\item[(a)]
We do not know whether the renorming assumption on $X$ in
Theorem~\ref{T:rozsir} can be omitted or essentially weakened.
\item[(b)]
The condition (ii) in Theorem~\ref{T:rozsir} can be substituted by the
following formally weaker condition:
\begin{enumerate}
\item[(ii')] {\em some control function of $F$ can be extended to a
d.c.\ function on $A$.}
\end{enumerate}
Indeed, if $f_1$ and $f_2$ are continuous convex functions on $A$ such
that $f_1-f_2$ controls $F$ on $C$, then also the
sum $f_1 +f_2$ controls $F$ on $C$.
\end{enumerate}
\end{remark}

\begin{corollary}\label{C:rozsir}
Let $X,Y$ be normed linear spaces, $C\subset X$ be a convex set with a
nonempty interior, and $F\colon C\to Y$ be a
d.c.\ mapping.
Supose that,
the restriction of $F$ to each bounded open convex subset of $C$ has a
Lipschitz control function.
\begin{enumerate}
\item[(a)] If $Y$ is a Banach space, then
           $F$ admits a d.c.\ extension to $\overline{C}$.
\item[(b)] If $X$ admits a renorming with modulus of convexity of
     power type 2, and either $C$ is closed or $Y$ is a Banach space, then
     $F$ admits a d.c.\ extension to the whole $X$.
\end{enumerate}
\end{corollary}

\begin{proof}
Consider the sets $A_n:=U(0,n)$ ($n\in\mathbb{N}$) and apply
Proposition~\ref{L:uzaver} to get (a), and Theorem~\ref{T:rozsir} to get
(b).
\end{proof}

\begin{corollary}\label{elpecka}
Let $X$ be a (subspace of some) $L_p(\mu)$ space with $1<p\le2$. Let
$C\subset X$ be a convex set with a nonempty interior.
\begin{enumerate}
\item[(a)] Each continuous convex function on $C$, which is Lipschitz on every
bounded subset of $\mathrm{int}\,C$, admits a d.c.\ extension to the whole $X$.
\item[(b)] Each Banach space-valued ${\mathcal C}^{1,1}$ mapping on $C$
admits a d.c.\ extension to the whole $X$.
\end{enumerate}
\end{corollary}

\begin{proof}
It is known (see e.g.\ \cite[p.410]{BeLin}) that $X$, in the $L_p$-norm,
has modulus of convexity of power type
2. Therefore, \cite[Proposition~1.11]{VeZa2} easily implies that
each Banach space-valued ${\mathcal C}^{1,1}$ mapping on any open
convex subset of $X$ is d.c.\ with a control function that is Lipschitz on
bounded sets. Now, both (a) and (b) follow from Corollary~\ref{C:rozsir}(b).
\end{proof}

For extensions from closed finite-dimensional convex subsets, we have
the following simple corollary.
Recall that a {\em finite-dimensional set} (in a vector space) is a set whose linear span is
finite-dimensional.

\begin{corollary}
Let $X,Y$ be normed linear spaces, and
$F\colon C\to Y$ be a d.c.\ mapping, where $C\subset X$ is a
finite-dimensional closed convex
set. Then the following assertions are equivalent:
\begin{enumerate}
\item[(i)] $F$ admits a d.c.\ extension $\widehat{F}\colon X\to Y$;
\item[(ii)] $F$ has a locally Lipschitz control function $f\colon C\to\mathbb{R}$.
\item[(iii)]For each $x \in C$, there exists $r_x >0$ such that the restriction of
$F$ to $C \cap U(x,r_x)$ has a
 Lipschitz control function.
\end{enumerate}
\end{corollary}

\begin{proof}
The implication $(i)\Rightarrow(ii)$ is obvious since each continuous convex
function on $X$ is locally Lipschitz. The implication $(ii)\Rightarrow(iii)$ is trivial.

Let $(iii)$ hold. Suppose that $0\in C$ and denote
$X_0:=\mathrm{span}\,C$ ($=\mathrm{aff}\,C$). Then $C$, being finite-dimensional,
has a nonempty interior in $X_0$.
 Let $B \subset C$ be a bounded convex set which is open in $X_0$. For each $x \in \overline{B} \cap C$ choose $r_x$
  by $(iii)$ and a Lipschitz convex function $\varphi_x$ on $C \cap U(x,r_x)$
which controls $F$ on $C \cap U(x,r_x)$. Since   $\overline{B} \cap C$ is compact, we can choose $x_1, \dots,x_n$ in
 $\overline{B} \cap C$ such that  $\overline{B} \cap C \subset \bigcup_{i=1}^n \, U(x_i, r_{x_i})$.
Extend $\varphi_{x_i}$ to a Lipschitz convex function $\psi_i$ on $X_0$
(cf. Lemma~\ref{F:1}(c)) and put
   $\psi := \sum_{i=1}^n \psi_i$. Then clearly $\psi|_{B}$ is a Lipschitz control function of $F|_B$.

By Corollary~\ref{C:rozsir}(b), there exists a d.c.\ extension
$F_0\colon X_0\to Y$ of $F$. Let $\pi\colon X\to X_0$ be a
continuous linear projection onto $X_0$. Then the mapping
$\widehat{F}:=F_0\circ\pi$ is a d.c.\ extension of $F$ (cf. \cite[Lemma 1.5(b)]{VeZa1}).
Thus (i) holds and the proof is complete.
\end{proof}

\section{Counterexamples}

\begin{example}\label{pasovec}
There exists a continuous convex function $f$ on the strip
$P:= \mathbb{R} \times [-1,0]$ such that
\begin{enumerate}
\item
$f$ has a d.c.\ extension to $\mathbb{R}^2$, and
\item
$f$ has no convex extension to $\mathbb{R}^2$.
\end{enumerate}
\end{example}

\begin{proof}
For  $(x,y) \in P$, we set
$$ f(x,y) := \sup \{a_t(x,y):\ t \in \mathbb{R}\},\  \text{where}\ \ a_t(x,y) := t^2 + 2t(x-t) + t^2 y.$$
Observe that
\begin{equation}\label{supp}
a_t(\cdot,0) \ \text{is the support affine function to the function}\ \ p(x):=x^2\ \text{at}\ t,
\end{equation}
\begin{equation}\label{nez}
a_t(t,y) \geq 0 \ \ \text{for}\ \ y \in [-1,0],\ \text{and}
\end{equation}
\begin{equation}\label{partder}
\frac{\partial a_t}{\partial x} (z) = 2t,\ \ \ \frac{\partial a_t}{\partial y} (z) = t^2\ \ (z\in \mathbb{R}^2).
\end{equation}
Now fix $\tau \in \mathbb{R}$ and consider a $t \in \mathbb{R}$. Then \eqref{supp} implies $a_t(\tau,0) \leq p(\tau) = \tau^2$,
 so \eqref{partder} gives $a_t(\tau,y) \leq \tau^2$ for $y \in [-1,0]$.
Consequently, $f(\tau,0) = a_{\tau}(\tau,0)= \tau^2$ and
  $f(\tau,y) \leq \tau^2 < \infty$
  for $y \in [-1,0]$. Thus $f$ is a finite convex function on $P$.
Moreover $f\ge0$ on $P$.

Note that $a_t(\tau,0) = -t^2 + 2t \tau \leq 0$ whenever $|t| \geq 2 |\tau|$.
If $z=(z_1,z_2)\in(\tau-1,\tau +1) \times [-1,0]$ and
$|t|\ge2(|\tau|+1)$, then $a_t(z_1,0)\le0$ since $|t|\ge2|z_1|$. For
such $z$ we have
$a_t(z)\le0\le f(z)$ because $a_t(z_1,\cdot)$ is nondecreasing by
\eqref{partder}. It follows that
  \begin{equation}\label{supmen}
  f(z) = \sup \{a_t(z):\ |t| \leq 2(|\tau|+1) \} \ \ \text{for}\ \ z \in (\tau-1,\tau +1) \times [-1,0].
  \end{equation}
  Using \eqref{supmen} and \eqref{partder}, we easily obtain that $f$ is locally Lipschitz on $P$; so it is Lipschitz
   on each bounded subset of $P$. Consequently,  (i) follows from
Corollary~\ref{C:rozsir}.

Now, suppose that (ii) is false, that is, there exists a
convex extension $\tilde{f}\colon\mathbb{R}^2\to\mathbb{R}$ of $f$.
Since $\tilde f(\tau,0) = f(\tau,0)=
    \tau^2$, we have  $\frac{\partial \tilde f}{\partial x} (\tau,0) = 2 \tau$ for each $\tau \in \mathbb{R}$.
    Now we will prove that, for each $\tau >0$,
    \begin{equation}\label{smder}
     d^+_{(0,-1)} \tilde f (\tau,0) =  d^+_{(0,-1)} a_{\tau} (\tau,0) = -\tau^2
    \end{equation}
where $d^+_v g(z)$ denotes the one-sided derivative of $g$ at $z$ in the
direction $v$.
     To this end, choose an arbitrary $\varepsilon >0$ and find $0<\delta< \sqrt{\varepsilon}$ such that $|t^2 - \tau^2| < \varepsilon$ whenever
      $|t-\tau|< \delta$. If $|t-\tau| \geq \delta$ and  $y \in (-\delta^2/\tau^2,0]$, then
      \begin{equation*}
\begin{split} a_{\tau}(\tau,y) - a_t(\tau,y)  &= \tau^2 + \tau^2y + t^2 - 2t\tau -t^2y \\
& = (t-\tau)^2 + (\tau^2-t^2)y \geq \delta^2 + \tau^2 y  > 0.
\end{split}
\end{equation*}
     If $|t-\tau|< \delta$ and $y\le0$, then
     $$ a_{\tau}(\tau,y) - a_t(\tau,y)   = (t-\tau)^2 + (\tau^2-t^2)y \geq \varepsilon y.$$
     Therefore,    $\tilde f(\tau,y) \geq a_{\tau}(\tau,y) \geq \tilde f(\tau,y) + \varepsilon y$
whenever $y \in (-\delta^2/\tau^2,0]$.
Since $\varepsilon>0$ was arbitrary,
we easily obtain \eqref{smder}.

Since $\tilde f$ is convex, the function $v \mapsto d^+_v \tilde f (\tau,0)$ is
positively homogenous and subadditive.
 Therefore, for $\tau>0$, we have
 $$ d^+_{(\tau,-3)} \tilde f (\tau,0) \leq  d^+_{(0,-3)} \tilde f (\tau,0) +
  d^+_{(\tau,0)} \tilde f (\tau,0) = -3\tau^2 + 2 \tau^2 = - \tau^2,$$
  and consequently  $d^+_{(-\tau,3)} \tilde f (\tau,0) \geq \tau^2$. By convexity of $\tilde f$,
  $$ \tilde f (0,3) \geq \tilde f(\tau,0) + d^+_{(-\tau,3)} \tilde f (\tau,0) \geq \tau^2 + \tau ^2 = 2\tau^2.$$
Since  $\tau >0$ was arbitrary, $\tilde f (0,3) = \infty$, a contradiction.
\end{proof}

\begin{example}\label{koulovec}
In $X=\ell_2$,
there exist a closed convex set
$C\subset U(0,1)$ with a nonempty interior
and a continuous convex function
$f\colon C\to\mathbb{R}$ such that:
\begin{enumerate}
\item[(a)] $f$ has a continuous convex extension to
$U(0,1)$, in particular, $f$ is locally Lipschitz on
$C$
(even there exists a nondecreasing sequence of open convex sets
$A_n\nearrow U(0,1)$ such that $f$ is Lipschitz on each
$A_n\cap C$);
\item[(b)] $f$ has no d.c.\ extension to  $U(0,r)$ whenever $r>1$.
\end{enumerate}
\end{example}

\begin{proof}
Let $e_n$ be the $n$-th vector of the standard basis of $X=\ell_2$.
For $n,k\in\mathbb{N}$ with $n<k$, put
$$\textstyle
z_{n,k}=(1-\frac1n)e_n + h_n(1-\frac1k)e_k
$$
where $h_n>0$ is such that $(1-\frac1n)^2+h_n^2=1$. Note that
$\|z_{n,k}\|^2=(1-h_n^2)+h_n^2(1-h_k^2)=1-h_n^2 h_k^2$. Put
$$\textstyle
C:=\overline{\mathrm{conv}}\left[
\frac12 B_{X}\cup\{z_{n,k}:\; n,k\in\mathbb{N},\;n<k\}
\right].
$$
Obviously, $C$ is a closed convex set with a nonempty interior and  $C \subset B_X$. We claim
that $C\subset U(0,1)$.

If this is not the case, there exists $x\in C$ with $\|x\|=1$. Thus
$\sup\langle x,C\rangle=\langle x,x\rangle=1.$ On the other hand,
there exists $n_0\in\mathbb{N}$ such
that
$|\langle x,e_n\rangle|<\frac13$ and $h_n<\frac13$ whenever $n>n_0$.
Thus $|\langle x,z_{n,k}\rangle|\le\frac23$ for $k>n>n_0$. There
exists
$k_0>n_0$ such that
$|\langle x,e_k\rangle|<\frac{1}{2n_0}$ whenever $k>k_0$. Hence, for
$n\le n_0$ and $k>k_0$, we have
$|\langle x,z_{n,k}\rangle|\le(1-\frac1n)+\frac{1}{2n_0}\le
1-\frac{1}{n_0}+\frac{1}{2n_0}=1-\frac{1}{2n_0}$.
Since obviously
$\sup\langle x,\frac12 B_{X}\rangle=\frac12$, we obtain
\begin{align*}
\sup\langle x,C\rangle&=\textstyle
\max\bigl\{\frac12 , \sup\left\{\langle x,z_{n,k}\rangle:\; n,k\in\mathbb{N},\;n<k\right\}\bigr\}\\
&\le\textstyle
\max\bigl[\{\frac23, 1-\frac{1}{2n_0}\} \cup
\{\|z_{n,k}\|:\; n<k\le k_0,\;n\le n_0\}\bigr]<1.
\end{align*}
This contradiction proves our claim.

The function $x\mapsto 1-\|x\|$ is positive, continuous and concave on
$U(0,1)$. Since the function $t\mapsto\frac1t$ is convex
and decreasing on $(0,\infty)$, the composed function
$g(x)=\frac{1}{1-\|x\|}$ is convex
continuous, and hence locally Lipschitz, on $U(0,1)$. Thus $f:=g|_C$
satisfies (a) by Lemma~\ref{K}. Let us show (b). By Lemma~\ref{ndc}, it
suffices to prove
that $g$ is unbounded on subsets of $C$
of arbitrarily
small diameter. Fix $n\in\mathbb{N}$. For any
two distinct indices $k,l>n$,  we have
$$\textstyle
\|z_{n,k}-z_{n,l}\|^2=
h_n^2\left[(1-\frac1k)^2+(1-\frac1l)^2\right]\le 2h_n^2
$$
which implies $\mathrm{diam}\,\{z_{n,k}: k>n\}\le\sqrt{2}\,h_n$.
This completes the proof since $g(z_{n,k})\to\infty$ as $k\to\infty$.
\end{proof}

\section{Extensions of convex functions from subspaces}

Let $Y$ be a closed subspace of a normed linear space $X$, and
$f\colon Y\to \mathbb{R}$ a continuous convex function. The present section
concerns the problem of existence of a continuous convex extension
$\hat{f}\colon X\to\mathbb{R}$ of $f$.

An example of nonexistence of $\hat f$ was given in \cite[Example~4.2]{BMV}.
On the other hand, it is easy and well known that such $\hat f$ exists if either $Y$ is
complemented in $X$ or $f$ is Lipschitz (see, e.g., \cite{BMV}). Borwein
and Vanderwerff proved in \cite[Fact, p.1801]{BV} that $\hat f$ exists
whenever $f$ is bounded on each bounded subset of $Y$; however, this
sufficient condition is not necessary (see Remark~\ref{nenutna}).
The following theorem contains a necessary and sufficient condition $(iv)$
of the same type, but the proof is more difficult and uses different
methods.
Our main new observation is that a modification of Hartman's construction
from \cite{H} gives the implication $(iv)\Rightarrow(ii)$; and we use also
the implication $(ii)\Rightarrow(i)$
already proved in \cite{BMV}.

\begin{theorem}\label{indiv}
Let $X$ be a normed linear space, $Y \subset X$ its closed subspace,
and $f\colon Y \to \mathbb{R}$ a continuous convex function.
 Then the following statements are equivalent.
 \begin{enumerate}
 \item
 The function $f$ admits a continuous convex extension to $X$.
 \item
 There exists a continuous convex function $g\colon X \to \mathbb{R}$ such that $f \leq g|_{Y}$.
 \item
 $f$ admits a d.c.\ extension to $X$.
 \item
 There exists a sequence $\{C_n\}$ of nonempty open convex subsets of $X$ such that $C_n \nearrow X$ and
  $f$ is bounded on each set $C_n \cap Y\ (n \in \mathbb{N})$.
 \item
 There exists a sequence $\{B_n\}$ of nonempty open convex subsets of $X$ such that $B_n \nearrow X$ and
  $f$ is Lipschitz on each set $B_n \cap Y\ (n \in \mathbb{N})$.
   \end{enumerate}
\end{theorem}
\begin{proof}
$(i)\Rightarrow(ii)$ is obvious, while $(ii)\Rightarrow(i)$ was proved
in \cite[Lemma~4.7]{BMV} for $X$ a Banach space. However, the proof
therein works also in the normed linear case (note that the convex
extension $\tilde{f}$ from \cite[Lemma~4.7]{BMV} is continuous since it
is locally upper bounded).

$(i)\Rightarrow(iii)$ is trivial.

$(iii)\Rightarrow(ii)$. If (iii) holds, there exist continuous convex
functions $u,v$ on $X$ such that $u(y)-v(y)=f(y)$ for $y\in Y$. Choose a
continuous affine function $a$ on $X$ such that $a\le v$. Then
$$
u(y)-a(y)=f(y)+(v(y)-a(y))\ge f(y),\quad y\in Y;
$$
so we can put $g:=u-a$.

$(i)\Rightarrow(v)$ follows immediately applying Lemma~\ref{K} to a
continuous convex extension $\hat{f}$ of $f$.

$(v)\Rightarrow(iv)$. Clearly, it suffices to put
$C_n=B_n\cap U(0,n)$ ($n\in\mathbb{N}$).

It remains to prove $(iv) \Rightarrow (ii)$.
Using Lemma~\ref{L:dn} and an obvious shift of indices,
it is easy to find a sequence $\{D_n\}$ of bounded open convex
subset of $X$ such that (for each $n$)
$D_n \cap Y \neq\emptyset$, $f$ is bounded on $D_n\cap Y$,
$d_n: = \mathrm{dist}(D_n, X \setminus D_{n+1})>0$, and $D_n \nearrow X$.

 Now we will construct inductively a sequence $(g_n)_{n\in \mathbb{N}}$ of
functions on $X$ such that, for each $n \in \mathbb{N}$,
 \begin{enumerate}
 \item[(a)]
 $g_n$ is convex and Lipschitz;
 \item[(b)]
 $g_n = g_{n-1}$ \ on\ $D_{n-1}$ whenever  $n>1$;
 \item[(c)]
 $g_n \geq f$ \ on \ $D_{n+1} \cap Y$.
\end{enumerate}
Set   $M_n := \sup\{f(y):\ y \in D_n \cap Y\}$; by the assumptions $M_n < \infty$.

Define $g_1(x) := M_2,\ x \in X$. Then the conditions (a), (b), (c) clearly hold for $n=1$.

Now suppose that $k>1$ and  we already have
$g_1,\dots, g_{k-1}$ such that  (a), (b), (c)  hold for each $1 \leq n <k$.
 We can clearly choose $a \in \mathbb{R}$ such that $g_{k-1}(x) \geq a$ for each $x \in D_{k-1}$, and then $b > 0$ such that
  $a + b\, d_{k-1} \geq M_{k+1}$. Define
  $$ g_k(x) := \max \{g_{k-1}(x), a + b\, \mathrm{dist} (x, D_{k-1})\},\ \ \ x \in X.$$
  We will show that  the conditions (a), (b), (c) hold for $n=k$.
   The validity of (a) is obvious. If $x \in D_{k-1}$, then $g_k(x) = \max\{g_{k-1}(x),a\} = g_{k-1}(x)$; so (b) holds.

   Now consider an arbitrary  $y \in D_{k+1} \cap Y$. If $y \in D_{k-1}$, using (b) for $n=k$ and (c) for $n=k-1$,
we obtain $g_k(y) = g_{k-1}(y) \geq f(y)$. If $y \in D_k \setminus D_{k-1}$, using the definition of $g_k$ and (c) for
    $n=k-1$, we also obtain $g_k(y) \ge g_{k-1}(y) \geq f(y)$. If $y \in D_{k+1} \setminus D_k$, then
    $$ g_k(y) \geq a + b \, \mathrm{dist}(y, D_{k-1}) \geq a + b\, d_{k-1} \geq M_{k+1} \geq f(y).$$
Now, for each $x \in X$, the sequence $\{g_n(x)\}$ is constant for large $n$'s, hence  $g(x):= \lim_{n\to \infty} g_n(x)$
 is defined on $X$. Since $g = g_n$ on $D_n$ by (b),
the conditions (a) and (c) easily imply that $g$ is a continuous
  convex function on $X$ such that $f \leq g|_{Y}$.
\end{proof}

\begin{remark}\label{nenutna}
As already mentioned, (i) holds whenever
\begin{enumerate}
\item[$(*)$] $f$ is bounded on each bounded subset of $Y$
\end{enumerate}
(indeed, (iv) holds with $C_n:=U(0,n)$). To see that $(*)$ is not
necessary with $Y\ne X$, consider an arbitrary infinite dimensional
Banach space $X$, a closed subspace $Y$ of finite codimension in $X$,
and a continuous convex function $f$ on $Y$ which is unbounded on some
bounded set (for its existence, see \cite{BFV}).
\end{remark}

 \begin{theorem}\label{univ}
Let $X$ be a normed linear space and $Y \subset X$ its closed subspace.
Then the following statements are equivalent.
\begin{enumerate}
\item
Each continuous convex function $f\colon Y\to\mathbb{R}$ admits a continuous convex extension to $X$.
\item
If $\{C_n\}$ is a sequence of open convex subsets of $Y$ such that
$C_n\nearrow Y$, then there exists
a sequence $\{D_n\}$ of open convex subsets of $X$ such that
$D_n\nearrow X$ and $D_n\cap Y\subset C_n$.
\item
If $\{C_n\}$ is a sequence of open convex subsets of $Y$ such that
$C_n\nearrow Y$, then there exists
a sequence $\{\tilde{C}_n\}$ of open convex subsets of $X$ such that
$\tilde{C}_n\nearrow X$ and $\tilde{C}_n\cap Y = C_n$.
\end{enumerate}
\end{theorem}

\begin{proof}
$(i)\Rightarrow(ii)$.
Let $\{C_n\}$ be as in (ii). Using Lemma \ref{L:dn}, we can (and do) suppose
that $C_n\ne\emptyset$ and $C_n\subset\subset C_{n+1}$ in $Y$ ($n\in\mathbb{N}$). Fix $a\in C_1$
and put $C_0:=\{a\}$. Choose $\epsilon_n>0$ such that
$C_n+\varepsilon_n B_Y\subset C_{n+1}$ ($n\ge0$), and consider the function
\[
f(y):=\sum_{n=0}^\infty \frac{1}{\varepsilon_n}\,\mathrm{dist}(y,C_n)\,,\quad
y\in Y.
\]
It is easy to see that $f$ is a continuous convex function on $Y$;
therefore it admits a continuous convex extension $\hat{f}$ to $X$ by
(i). Let us show that the sets $D_n:=\{x\in X: \hat{f}(x)<n\}$ ($n\in\mathbb{N}$) have the
desired properties. Obviously, they are convex and open, and $D_n\nearrow
X$. Consider $n\in\mathbb{N}$ and $y\in Y\setminus C_n$. Since
$\mathrm{dist}(y,C_k)\ge \epsilon_k$ for $0\le k<n$, we have
$$
f(y)\ge\sum_{k=0}^{n-1}\frac{1}{\epsilon_k}\,\mathrm{dist}(y,C_k)\,\ge n.
$$
This shows that $D_n\cap Y\subset C_n$.

$(ii)\Rightarrow(i)$.
Let $f$ be as in (i). Then the sets
$C_n:=\{y\in Y: f(y)<n,\;\|y\|<n\}$ ($n\in\mathbb{N}$) are open convex and satisfy
$C_n\nearrow Y$. Observe that $f$ is bounded on each $C_n$ by
Lemma~\ref{F:1}(a).
Find $D_n$ ($n\in\mathbb{N}$) by (ii). Since $D_n\cap Y\subset C_n$, the sequence
$\{D_n\}$ satisfies the condition (iv) of Theorem~\ref{indiv}, and so (i)
follows.

$(ii)\Rightarrow(iii)$.
Let $\{C_n\}$ be as in (iii). Find $D_n$ ($n\in\mathbb{N}$) by (ii). Choose
$n_0\in\mathbb{N}$ such that $D_{n_0}\cap Y\ne\emptyset$. For $n\ge n_0$, put
$\tilde{C}_n:=\mathrm{conv}(D_n\cup C_n)$. By Lemma~\ref{conv}(a),(c),
we have that $\tilde{C}_n\cap Y=C_n$ and the convex set $\tilde{C}_n$ is open
for any $n\ge n_0$. Let $n_1$ be the smallest index such that
$C_{n_1}\ne\emptyset$.
Fix $c\in C_{n_1}$ and choose $r>0$ such that
$U(c,r)\subset\tilde{C}_{n_0}$ and
$U(c,r)\cap Y\subset C_{n_1}$. Put $\tilde{C}_n=\emptyset$ for
$1\le n<n_1$, and $\tilde{C}_n:=\mathrm{conv}(U(c,r)\cup C_n)$ for
$n_1\le n<n_0$. Using Lemma~\ref{conv}(a),(c) as above, we easily obtain
that the sequence $\{\tilde{C}_n\}$ has the desired properties.
The reverse implication $(iii)\Rightarrow(ii)$ is obvious.
\end{proof}

As an application of Theorem~\ref{univ}, we give an alternative proof
(see Theorem~\ref{subspace}) of the fact that separability of the quotient
space $X/Y$ is sufficient for extendability of all continuous convex
functions on $Y$ to $X$. This was proved in \cite{BMV} for Banach spaces
using a condition about nets in $Y^*$, equivalent to
(i) of Theorem \ref{univ}, together with  Rosenthal's extension theorem.
Our proof (for general normed
linear spaces) is based on Theorem~\ref{univ} and on the following
elementary lemma.

\begin{lemma}\label{kuzeliky}
Let $Y$ be a closed subspace of a normed linear space $X$. Let $B=r B_X$ for some
$r>0$. Then, for any $x\in X$, there exists $y_x\in Y$ such that
\begin{equation}\label{E:kuzeliky}
\mathrm{conv}[(x+B)\cup B] \cap Y
\subset \mathrm{conv}[\{y_x\}\cup 8B].
\end{equation}
\end{lemma}

\begin{proof}
If $x=0$, $y_x=0$ works. For
$x\ne0$,
denote $P:=\mathrm{conv}[(x+B)\cup B]\cap Y$ ($\ne\emptyset$) and
$s:=\sup\{\|y\|:y\in P\}$, and
choose $u_0\in P$ such that $\|u_0\|>s-r$.
Observe that $u_0\ne0$ since $s\ge\sup\{\|y\|:y\in B\cap Y\}=r$. We claim
that $y_x=8u_0$ works.

Fix $a_0\in x+B$, $b_0\in B$ and $\lambda\in[0,1]$ such that
$u_0=(1-\lambda)a_0+\lambda b_0$.
It suffices to prove that $P\setminus B\subset \mathrm{conv}[\{y_x\}\cup 8B].$
Given $u\in P\setminus B$, choose $a\in x+B$ and $b\in B$ such that
$u\in[a,b]$. Note that $a\ne b$ since $u\notin B$.
Put $v=(1-\lambda)a+\lambda b$ and observe that $\|v-u_0\|\le 2r$.
Consider the half-line
$H:=\{v+t(a-b): t\ge0\}$. Let $v_1\in H$ be such that $\|v_1-v\|=5r$.
Then $v_1\in u_0+7B$ since
$\|v_1-u_0\|\le\|v_1-v\|+\|v-u_0\|\le 7r$.

We claim that no point $y\in H$ with $\|y-v\|>5r$ can belong to $P$ since it satisfies
$\|y\|>s$.
Indeed, since $v\in[y,b]$,
\begin{align*}
\|y\| &\ge\|y-b\|-\|b\|=\|y-v\|+\|v-b\|-\|b\|
\ge \|y-v\| +\|v\|-2\|b\| \\
& \ge \|y-v\| +\|u_0\|-\|u_0-v\|-2\|b\| > 5r+(s-r)-2r-2r=s.
\end{align*}
Consequently, if $u\in H$ then $u\in[v_1,v]$,
and if $u\in[a,b]\setminus H$ then $u\in[v,b]$. In both cases,
$u\in[v_1,b]\subset\mathrm{conv}[(u_0+7B)\cup B]$.
To finish, observe that $u_0+7B=\frac18(8u_0)+\frac78(8B)$ implies
$$
u\in \mathrm{conv}[(u_0+7B)\cup B]\subset
\mathrm{conv}[\{8u_0\}\cup 8B].
$$
\end{proof}

\begin{theorem}[{\cite[Corollary~4.10]{BMV}} for $X$ Banach]\label{subspace}
Let $Y$ be a closed subspace of a normed linear space $X$ such
that $X/Y$ is separable. Then each continuous convex function
$f\colon Y\to\mathbb{R}$ admits a continous convex extension to $X$.
\end{theorem}

\begin{proof}
It suffices to verify the condition (ii) of
Theorem~\ref{univ}.
Let $C_1\subset C_2\subset\ldots$ be
open convex subsets of $Y$ such that
$\bigcup_n C_n=Y$.
We can (and do) suppose that
$0\in\mathrm{int}_Y\,C_1$. Fix $r>0$ such that
\begin{equation}\label{E:8r}
8 r B_Y\subset C_1.
\end{equation}
Fix a dense sequence $\{\xi_n\}_{n\in\mathbb{N}}\subset X/Y$ and,
for each $n$, choose an arbitrary $z_n\in\xi_n$. The sets
$Z_n:=\mathrm{conv}\{z_1,\ldots,z_n\}$ ($n\in\mathbb{N}$) form a nondecreasing
sequence of compact convex sets such that the union
$\bigcup_n(Z_n+Y)$ is dense in $X$. Define $Z_0=\emptyset$.

{\it Claim.} There exists an increasing sequence of
integers $\{k_n\}_{n\ge0}$ such that $k_0=1$ and, for each $n$,
\begin{equation}\label{E:claim}
\mathrm{conv}(Z_n\cup B)\cap Y\subset C_{k_n}\qquad
\text{where $B=rB_X$.}
\end{equation}
To prove this, we shall proceed by induction with respect to $n$.
Observe that \eqref{E:claim} is satisfied for $n=0$ and $k_0=1$.
Suppose we already have
$k_0,\ldots,k_{n-1}$. Since $Z_n$ is compact, there exists a finite set
$F\subset Z_n$ such that $Z_n\subset F+B$. For any $x\in F$, fix $y_x\in Y$
satisfying \eqref{E:kuzeliky}. Choose an integer $k_n>k_{n-1}$ such that
$y_x\in C_{k_n}$ for each $x\in F$. Then, using \eqref{Val}, we obtain
\[
\mathrm{conv}(Z_n\cup B)=
\bigcup_{z\in Z_n}\mathrm{conv}(\{z\}\cup B)\subset
\bigcup_{x\in F}\mathrm{conv}((x+B)\cup B).
\]
Consequently, using \eqref{E:kuzeliky} and Lemma \ref{conv}(a), we obtain
\begin{align*}
\mathrm{conv}(Z_n\cup B)\cap Y &\subset
\bigcup_{x\in F}[\mathrm{conv}((x+B)\cup B)\cap Y] \\
&\subset
\bigcup_{x\in F}[\mathrm{conv}(\{y_x\}\cup 8B)\cap Y] \\
&=
\bigcup_{x\in F} \mathrm{conv}[(8B\cap Y)\cup \{y_x\}] \subset C_{k_n}
\end{align*}
since, by \eqref{E:8r}, $(8B\cap Y)\cup \{y_x\} \subset C_{k_n}$ for each $x\in F$.
This proves our Claim.

For each $j\in\mathbb{N}$, let $n(j)$ be the unique nonnegative integer
with $k_{n(j)}\le j <k_{n(j)+1}$.
Let us define a nondecreasing sequence $\{D_j\}_{j\in\mathbb{N}}$ of
open convex sets by
$$
D_j:=\mathrm{int}[\mathrm{conv}(Z_{n(j)}\cup B\cup C_j)].
$$
By Lemma~\ref{conv}(a) and \eqref{E:claim}, we have
\begin{align*}
Y\cap D_j&\subset
Y\cap \mathrm{conv}(Z_{n(j)}\cup B\cup C_j)\\
&=
Y\cap\mathrm{conv}[\mathrm{conv}(Z_{n(j)}\cup B)\cup C_j]
\\&=
\mathrm{conv}\bigl\{[Y\cap\mathrm{conv}(Z_{n(j)}\cup B)]\cup C_j\bigr\}
\\&\subset
\mathrm{conv}\{C_{k_{n(j)}}\cup C_j\}=C_j.
\end{align*}
It remains to prove that $\bigcup_j D_j=X$. By Lemma~\ref{conv}(b), this
is equivalent to say
that $\bigcup_j D_j$ is dense in $X$. Since, for each $j$, $D_j$ is
dense in $\tilde{D}_j:=\mathrm{conv}(Z_{n(j)}\cup B\cup C_j)$, it
suffices to show that $\bigcup_j \tilde{D}_j$ is dense.
Note that $H:=\frac12\,\bigcup_n(Z_n+Y)$ is dense in $X$ since
$\bigcup_n(Z_n+Y)$ is dense.
If $h\in H$ then
$h=\frac12(z+y)$ with $z\in Z_n$ for some $n$, and $y\in Y$. Then, for
sufficiently large $j$, we have $z\in Z_{n(j)}$ and $y\in C_j$, and
hence $h\in \tilde{D}_j$. Consequently,  $\bigcup_j \tilde{D}_j$
is dense since it
contains $H$.
\end{proof}

\bigskip
\bigskip

\subsection*{Acknowledgement}
The research of the first author was partially supported by the Ministero
dell'Istruzione, dell'Universit\`a e della Ricerca
 of Italy.
The research of the second author was partially supported by the grant
GA\v CR 201/06/0198  from the Grant Agency of
Czech Republic and partially supported by the grant MSM  0021620839 from
 the Czech Ministry of Education.