Let $A$ be a $C^\*$-algebra and $\pi:A\rightarrow B(H)$ a faithful $\*$-representation, so $M=\pi(A)''$ is a von Neumann algebra and $A\rightarrow M$ is an inclusion. von Neumann's Bicommutant Theorem tells us that $A=\pi(A)$ is weak$^\*$-dense in $M$, and the Kaplansky Density Theorem says that further, the unit ball of $A$ is weak$^\*$-dense in that of $M$.


Suppose now I have $a\in A$ and $k\geq 0$ fixed, and there is $x\in M$ with
$$ \|a-x\|\leq k, \quad \|x\|\leq 1. $$



> 
> Is $x$ is the weak$^\*$-closure of the set $\{ b\in A : \|a-b\|\leq k, 
> \|b\|\leq 1 \}$? 
> 
> 
> 


If $a=0$ this is just Kaplansky density.


Let's weaken this, and just ask: is there $b\in A$ with $\|a-b\|\leq k$ and $\|b\|\leq 1$? This follows from just the triangle-inequality, because it is easy to see that
$$ \inf\{k\geq 0 : \exists b, \|a-b\|\leq k, \|b\|\leq 1\} = \max(0, \|a\|-1). $$
So if $\|a-x\|\leq k$ and $\|x\|\leq 1$ then $\|a\|\leq k+1$ and so $\|a\|-1\leq k$.


So, let's make this new problem harder. Let $a\_0\in A^+$ (so $a\_0$ is positive: this is motivated by other considerations) and ask the following:



> 
> Suppose there is $x\in M$ with $\|a-a\_0x\|\leq k$ and $\|x\|\leq 1$. Is there $b\in A$ with $\|a-a\_0b\|\leq k$ and $\|b\|\leq 1$?
> 
> 
> 


One could also consider more general maps $T:A\rightarrow A$ which extend to $M\rightarrow M$; here $T(b) = a\_0b$.



Variations on Kaplansky Density
The answer to the first question is YES (~~and probably the second as well~~ ). 


First of all, we may replace the original representation $\pi(A)\subset B(H)$ with a universal representation $\pi\oplus\sigma$, by replacing $x \in \pi(A)''$ with $x\oplus \frac{1}{k+1}\sigma(a) \in A^{\*\*}$. 
Then, Kaplansky's density theorem is upgraded to the following (which is 
an easy consequence of the Hahn--Banach separation theorem).


Lemma 1: Let $z \in A^{\*\*}$, $w \in A$, and a net $(z\_i)\_i$ in $A$. 
If $\| z - w \| \le 1$ and $z\_i \to z$ weak\*, then 
$$\lim\_j\mathrm{dist}(w,\mathrm{conv}\{ z\_i : i \geq j\})\le1$$


The advantage of using convex combination is that it can be iterated 
without destroying the previously obtained approximation estimate.
From Lemma 1, one immediately obtains 


Lemma 2: Let $x\in A^{\*\*}$ and $a\in A$ be such that 
$\|x\|\le1$ and $\| x - a \|\le k$. Then for any $\epsilon\_1>0$, 
the element $x$ is weak\*-approximated by $y\_1\in A$ such that 
$\|y\_1\|\le 1+\epsilon\_1$ and $\| y\_1 - a \|\le k+\epsilon\_1$. 


We are done once we show the approximant $y\_1$ in Lemma 2 is norm-close 
to an element that satisfies the exact norm inequalities: 


Lemma 3: Let $y\_1 \in A$ and $a \in A$ be such that $\|y\_1\|\le1+\epsilon\_1$, $\|y\_1-a\|\le k+\epsilon\_1$, and $\|a\|\le k+1$. 
Then, there is $y\in A$ such that $\|y\|\le1$, $\|y-a\|\le k$, and $\|y-y\_1\|\approx\_{\epsilon\_1}0$. 
Here $\approx\_{\epsilon\_1}$ means that the difference is at most 
$h(\epsilon\_1)$ for some explicit continuous function $h\geq0$ 
such that $h(0)=0$. 


Now Lemma 3 is proved by iterating the following approximate version and finding a suitable convergence sequence $(y\_n)\_n$: 


Lemma 4: Let $y\_1 \in A$ and $a \in A$ be such that $\|y\_1\|\le1+\epsilon\_1$, $\|y\_1-a\|\le k+\epsilon\_1$, and $\|a\|\le k+1$. 
Then, for any $\epsilon\_2>0$, there is $y\_2\in A$ such that $\|y\_2\|\le1+\epsilon\_2$, $\|y\_2-a\|\le k+\epsilon\_2$, and $\|y\_2-y\_1\|\approx\_{\epsilon\_1}0$. 


Proof of Lemma 4: By Lemma 1, it suffices to find $y\_2$ in $A^{\*\*}$ 
(as opposed to in $A$). 
Put $\alpha=\beta=(2\epsilon\_1)^{1/2}\approx\_{\epsilon\_1}0$. 
Let $a=v|a|$ be the polar decomposition,  

$p:=1\_{[k+1-\alpha,k+1]}(|a|)$, and $q:=vpv^\*$. 
Since $ap\approx\_{\epsilon\_1}(k+1)vp$, 
$\|y\_1p\|\approx\_{\epsilon\_1}1$, $\| y\_1p - ap \|\approx\_{\epsilon\_1}k$, 
and $vp$ is a partial isometry, 
one has $y\_1p \approx\_{\epsilon\_1}vp$ and $y\_1p \approx\_{\epsilon\_1} qy\_1$.
Thus for $a':=ap^\perp=q^\perp a p^\perp$ (which has $\|a'\|\le k+1-\alpha$) and 
$$y\_2:= qvp + q^\perp((1-\beta)\frac{y\_1}{\|y\_1\|}+\beta\frac{a'}{\|a'\|})p^\perp$$
one has 
$\| y\_2 \|\le 1$ and $y\_2\approx\_{\epsilon\_1}y\_1$. 
Moreover, since 
\begin{align\*}
\|q^\perp(y\_2-a)p^\perp\|&\le\|y\_1-\frac{y\_1}{\|y\_1\|}\|+\|(1-\beta)q^\perp y\_1p^\perp+\frac{\beta}{\|a'\|}a' - a'\|\\
&\le \epsilon\_1+(1-\beta)(k+\epsilon\_1)+\beta(\|a'\|-1)\\
&\le k+2\epsilon\_1-\alpha\beta = k,
\end{align\*}
one has $\|y\_2-a\|\le k$ (assuming $k>\alpha$). 


