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, $$ \|A\|_2^2 \leq \|A \|_p \|A \|_q\;\; \forall\, \frac1p + \frac1q = 1, \text{ or,} $$

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


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*}

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.