While perusing the [Matrix norms section of Wikipedia](https://en.wikipedia.org/wiki/Matrix_norm#Examples_of_norm_equivalence), I came across this generalized version of Holder's inequality.


$$
\|A\|\_2^2 \leq \|A \|\_1 \|A \|\_\infty\,,
$$


where,
$$
\|A \|\_p = \max\_{\|x\|\_p = 1} \|Ax\|\_p\,,
$$
is the subordinate norm.


I tried looking up the references mentioned in the wiki page, but couldn't find anything relevant. Does this always hold true? Are these more general inequalities also true,
$$
\|A\|\_2^2 \leq \|A \|\_p \|A \|\_q\;\; \forall\, \frac1p + \frac1q = 1, \text{ or,}
$$


$$
\|A B\|\_2 \leq \|A \|\_p \|B \|\_q\;\; \forall\, \frac1p + \frac1q = 1\,?
$$



Generalized Hölder's inequality for operator (subordinate) norms
Actually, there is a much stronger result, known as the **Riesz-Thorin Theorem**:



> 
> The subordinate norm $\|A\|\_p$ is a log-convex function of $\frac1p$.
> 
> 
> 


In other words,
$$\left(\frac1r=\frac\theta{p}+\frac{1-\theta}q\right)\Longrightarrow(\|A\|\_r\le\|A\|\_p^\theta\|A\|\_q^{1-\theta}).$$
This contains as a particular case the inequality $\|A\|\_2^2\le\|A\|\_p\|A\|\_{p'}$ that you mention. For a proof of R.-T. Theorem, see Section 7.3 (in the second edition) of my book *Matrices*, GTM**216**, Springer-Verlag (2010).


As for the last inequality, for $\|AB\|\_2$, it is deadly false. If it was correct, then taking $B=A^T$, one would have
$$\|A\|\_2^2=\|AA^T\|\_2\le\|A\|\_p\|A^T\|\_{p'}=\|A\|\_p^2,$$
hence $\|A\|\_2\le\|A\|\_p$ for every $p$ and $A$, which is obviously false. Indeed, take a rank-one matrix $A=xy^T$ ; then
$$\|A\|\_p=\|x\|\_p\|y\|\_{p'}.$$
Take two vectors $x$ and $y$ such that $\|x\|\_2=\|x\|\_1$ and $\|y\|\_2>\|y\|\_\infty$ (possible if $n\ge2$), then $\|A\|\_2 = \|x\|\_2\|y\|\_2> \|x\|\_1\|y\|\_\infty = \|A\|\_1$.


**Edit.** Here is the elementary proof of $\|A\|\_p\le\|A\|\_1^{1/p}\|A\|\_\infty^{1/p'}$. Recall the formulae
$$\|A\|\_1=\max\_j\sum\_i|a\_{ij}|,\qquad\|A\|\_\infty=\max\_i\sum\_j|a\_{ij}|.$$
Applying Hölder, one has
\begin{align}
\|Ax\|\_p^p & = \sum\_i\left|\sum\_ja\_{ij}x\_j\right|^p\\&\le\sum\_i\left(\sum\_j|a\_{ij}|\right)^{p/p'}\sum\_k|a\_{ik}|\,|x\_k|^p \\
& \le \|A\|\_\infty^{p-1}\sum\_{i,k}|a\_{ik}|\,|x\_k|^p\\
&=\|A\|\_\infty^{p-1}\sum\_k|x\_k|^p\sum\_i|a\_{ik}|\\
&\le\|A\|\_\infty^{p-1}\|A\|\_1\|x\|\_p^p.
\end{align}


