Let $p>2$. I'd like to know the best possible lower and upper bound for $\|x\|\_p$ given that $x\in R^n$ and $\|x\|\_1$, $\|x\|\_2$, and $\|x\|\_\infty$ have fixed values.


It is well-known that 
$$\|x\|\_\infty\le \|x\|\_p\le \|x\|\_2^{2/p}\|x\|\_\infty^{1-2/p}~~~
\Big[\le \|x\|\_2\le \|x\|\_1\Big].$$
But these bounds cannot be sharp when an arbitrary value of $\|x\|\_1$ is specified. 


In case no closed form solution is known, I'd also appreciate pointers to explicit suboptimal bounds that are stronger than the stated ones. For example, I conjecture that a bound of the form
$$ \|x\|\_p\le C\_p\|x\|\_2\Big(\frac{\|x\|\_\infty}{\|x\|\_1}\Big)^r$$
should hold with $r=(p-2)/(2p-2)$ and a constant $C\_p$ depending on $p$ only, but I don't have a proof of such an inequality.



norm inequalities
By rescaling, without loss of generality (wlog) $\|x\|\_\infty=1$. 
Let $X$ be a random variable such that $P(X=|x\_i|)=1/n$ for $x=(x\_1,\dots,x\_n)$ and all $i=1,\dots,n$. Then 
\begin{equation\*}
 \|x\|\_r^r=nm\_r,\quad m\_r:=EX^r, 
\end{equation\*}
for $r\ge1$. Let us find the best possible bounds on $m\_p$ in terms of $m\_0=1,m\_1,m\_2$; these bounds can then be easily translated back in the terms of $\|x\|\_r$. Wlog 
\begin{equation\*}
0<m\_1<\sqrt{m\_2}<\sqrt{m\_1}<1. \tag{1} 
\end{equation\*}
Everywhere here $0\le t,u\le1$. 



> 
> **Lemma 1.** $d\_1(t):=(p-1) u^{p-2}t^2+(2-p) u^{p-1}t-t^p\le0$, and $d\_1(t)=0$ if $t\in\{0,u\}$. 
> 
> 
> **Lemma 2.** $d\_2(t):=c t^2+b t+a-t^p\ge0$, and $d\_2(t)=0$ if $t\in\{u,1\}$, where 
>  \begin{align\*}
>  a&:=\frac{(p-2) u^{p+1}-(p-1) u^p+u^2}{(1-u)^2}, \\
>  b&:=\frac{p u^{p-1}-(p-2) u^{p+1}-2 u}{(1-u)^2}, \\ 
>  c&:=\frac{-p u^{p-1}+(p-1) u^p+1}{(1-u)^2}. 
> \end{align\*}
> 
> 
> 


These lemmas will be proved at the end of this answer. 


It follows from Lemma 1 that 
\begin{equation\*}
 0\ge Ed\_1(X)=(p-1) u^{p-2}m\_2+(2-p) u^{p-1}m\_1-m\_p, \tag{2}
\end{equation\*}
and this inequality turns into the equality if $u=u\_\*:=m\_2/m\_1$ and $P(X=u\_\*)=m\_1^2/m\_2=1-P(X=0)$; in view of (1), such a r.v. $X$ exists, and $u\_\*\in(0,1)$. So, substituting $u=u\_\*$ into (2), we have the best possible lower bound on $m\_p$: 
\begin{equation}
 m\_p\ge m\_1(m\_2/m\_1)^{p-1}. 
\end{equation}


Similarly, it 
It follows from Lemma 2 that 
\begin{equation\*}
 0\le Ed\_2(X)=c m\_2+b m\_1+a-m\_p, \tag{3}
\end{equation\*}
and this inequality turns into the equality if $u=u\_{\*\*}:=(m\_1-m\_2)/(1-m\_1)$ and $P(X=u\_{\*\*})=(1-m\_1)/(1-u\_{\*\*})=1-P(X=1)$; in view of (1), such a r.v. $X$ exists, and $u\_{\*\*}\in(0,1)$. So, substituting $u=u\_{\*\*}$ into (3), we have the best possible upper bound on $m\_p$: 
\begin{equation}
 m\_p\le \frac{m\_2-m\_1^2+(1-m\_1)^2 (\frac{m\_1-m\_2}{1-m\_1})^p}{1+m\_2-2 m\_1}. 
\end{equation}




---


It remains to prove Lemmas 1 and 2. 


*Proof of Lemma 1.* We have $d\_1(0)=0=d\_1(u)=d'\_1(u)$. Also, $d''\_1(t)=(p-1) (2 u^{p-2} - p t^{p-2})$ switches in sign from $+$ to $-$ (only once) as $t$ increases from $0$ to $1$, and the switch point is $<u$. Now Lemma 1 follows. $\Box$


*Proof of Lemma 2.* We have 
\begin{align\*}
 d''\_2(t)(1-u)^2u &=2 \left((p-1) u^{p+1}-p u^p+u\right)-(p-1) p (1-u)^2 u t^{p-2} \\ 
&>2 \left((p-1) u^{p+1}-p u^p+u\right)-(p-1) p (1-u)^2 u^{p-1} \\
 &=:f(u) 
\end{align\*}
if $0<t<u$. 
We have 
\begin{equation}
 f''(u)=(p-2) (p-1) p (1-u) u^{p-3} ((p+1)u-(p-1)), 
\end{equation}
so that $f''(u)$ switches in sign from $-$ to $+$ (only once) as $u$ increases from $0$ to $1$. Also, $f(0)=f(1)=f'(1)=0$. So, $f>0$ and hence $d''\_2(t)>0$ if $0<t<u$. Also, clearly $d\_2$ has at most one inflection point. Also, $d\_2(1)=0=d\_2(u)=d'\_2(u)$. Now Lemma 2 follows. $\Box$




---




---


*Remark 1.* The lower bound on $\|x\|\_p$ or, equivalently, on $m\_p$, is an easy corollary of Hölder's inequality or, equivalently, of the log-convexity of $\|x\|\_r$ in $r$. However, in this case, it seemed to make sense to use a similar approach to both the lower bound and (the much more difficult) upper bound. 


*Remark 2.* The bounds in terms of $m\_r$ seem more natural than those in terms of $\|x\|\_r$ (even though these two kinds of bounds are very easy to translate into each other). One reason for this preference is that restrictions (1) when translated into the $\|x\|\_r$ terms will contain $n$. Moreover, the restrictions
$m\_1<\sqrt{m\_2}$ and $\sqrt{m\_1}<1$ in (1), when translated into the $\|x\|\_r$ terms, become rather restrictions on $n$: $n>\|x\|\_1^2/\|x\|\_2^2$ and $n>\|x\|\_1$, respectively. In these remarks, it is still assumed that $\|x\|\_\infty=1$. 


*Remark 3.* The lower bound on $m\_p$, when translated into the $\|x\|\_r$ terms, is free of $n$: 
\begin{equation}
 \|x\|\_p^p\ge\|x\|\_2^{2p-2}/\|x\|\_1^{p-2}. 
\end{equation}


However, if one insists on a free of $n$ upper bound on $\|x\|\_p$ in terms of $p$, $\|x\|\_1$, and $\|x\|\_2$ only, then the bound cannot be substantially better than the trivial upper bound given by the inequality $\|x\|\_p^p\le\|x\|\_2^2$, as in the first display in the OP. Indeed, we have 



> 
> **Proposition 1.** Suppose that 
>  \begin{equation}
>  \|x\|\_p\le F(\|x\|\_1,\|x\|\_2^2)
> \end{equation}
>  for all vectors $x$, 
>  where the function $F=F\_p$ is continuous (and does not depend on $n$). Then 
>  \begin{equation}
>  F(A,B)\ge B^{1/p} 
> \end{equation}
>  for all natural $B$ and all real $A>B$. 
> 
> 
> 


The restriction $A>B$ corresponds to the restriction $\sqrt{m\_2}<\sqrt{m\_1}$ in (1). 


*Proof of Proposition 1.* Take indeed any natural $B$ and any real $A>B$. For natural $n>A$, let $y=(y\_1,\dots,y\_n)$, where $y\_1=\dots=y\_B=1$ and $y\_{B+1}=\dots=y\_n=\frac{A-B}{n-B}[\in(0,1)]$. Then $\|y\|\_1=A$, $\|y\|\_2^2\to B$, $\|y\|\_p^p\to B$, and hence 
\begin{equation}
 B^{1/p}\leftarrow\|y\|\_p\le F(\|y\|\_1,\|y\|\_2^2)\to F(A,B) 
\end{equation}
as $n\to\infty$. $\Box$


In particular, it follows from Proposition 1 that the conjecture 
$$ \|x\|\_p\le C\_p\|x\|\_2\Big(\frac{\|x\|\_\infty}{\|x\|\_1}\Big)^r$$
with $r=(p-2)/(2p-2)$, 
proposed by the OP, is false in general. This can also be seen directly by letting $x\_i=1$ for all $i=1,\dots,n$. 


