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While perusing the Matrix norms section of Wikipedia, 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, $$ \|A\|_2^2 \leq \|A \|_p \|A \|_q\;\; \forall\, \frac1p + \frac1q = 1, \text{ or,} $$

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

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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, GTM216, 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{eqnarray*} \|Ax\|_p^p & = & \sum_i\left|\sum_ja_{ij}\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{eqnarray*}

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  • $\begingroup$ Thanks @Denis! Great answer. I have a follow up question. I know that Holder's inequality is proved using Young's inequality, which is involves convexity. But with bit of algebraic manipulation, we can trivially prove that following for vector norms. $\|v\|_2^2 \leq \|v\|_1 \|v\|_\infty$. Is there a similar algebraic way of proving $\|A\|_2^2 \leq \|A\|_1 \|A\|_\infty$, in case of matrix norms? $\endgroup$ – GraspIt Mar 30 '17 at 14:35
  • $\begingroup$ @Grasplt. Proving the matrix norm inequality (Riesz-Thorin) is much harder, except in the cases where $(p,q)=(1,\infty)$. It involves complex variable and the use of the Hadamard's three-lines Lemma. $\endgroup$ – Denis Serre Mar 30 '17 at 14:40
  • $\begingroup$ Did you mean the case of $(p,q) = (1,\infty)$ involves Hadamard's three-lines Lemma? If not, could you provide some hints on how to prove this particular case or point me to some references where I can look it up? Thanks. $\endgroup$ – GraspIt Mar 30 '17 at 15:34
  • $\begingroup$ @Grasplt. The elementary case is with the pair $(1,\infty)$. See my edits. $\endgroup$ – Denis Serre Mar 30 '17 at 16:07

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