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I was wondering if the Hölder's inequality was true for matrix induced norms, i.e. if $$\|AB\|_1 \leq \|A\|_p\|B\|_q, \quad\forall p,q \in [1,\infty] \text{ s.t. } \tfrac{1}{p}+\tfrac{1}{q} = 1.$$ But it seems that this does not hold in general, in fact $$A = \begin{bmatrix}1 & 2\\ 0 & 0 \end{bmatrix}, \; B = \begin{bmatrix}1 & 0\\ 2 & 0 \end{bmatrix}, \; p = 1, \; q = \infty $$ is a simple counterexample, and it is not hard to find similar ones for other choices of $p$ and $q$.

Question Does a Hölder-like inequality hold for matrix induced norms?

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    $\begingroup$ Maybe a good rewording/related question is "Holder's inequality for function composition", i.e. take Holder's inequality and replace $fg$ with $f \circ g$. $\endgroup$ – usul Feb 27 '14 at 21:46
  • $\begingroup$ Thank you for the observation, @usul! Anyway, I don't know if an Hölder-like inequality for functions w.r.t. composition could imply something about my question in a straightforward way, because the $p$-norm of a function has a very different nature from the $p$-norm of a matrix... $\endgroup$ – Paglia Feb 27 '14 at 21:57
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    $\begingroup$ @usul I don't see why this would be a good idea. The question asks about an inequality involving the norm of a product of elements of an algebra (which is what H\"older's inequality also treats) and so composition of scalar-valued functions does not seem obviously relevant. $\endgroup$ – Yemon Choi Feb 28 '14 at 2:15
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    $\begingroup$ @Yemon,Paglia: agreed that it is not all that related, I just reacted without stopping to think about operator norms vs function norms and so on. $\endgroup$ – usul Feb 28 '14 at 2:36
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There are (at least two) "generalizations" of Hölder inequality to the non-commutative case. One is the so called tracial matrix Hölder inequality:

$$ |\langle A, B \rangle_{HS} |= |\mathrm{Tr} (A^\dagger B) | \le \| A\|_p \,\, \| B\|_q $$

where $\| A\|_p$ is the Schatten $p$-norm and $1/p+1/q=1$. You can find a proof here.

Another generalization is very similar to what you wrote and reads

$$ \parallel|AB|\parallel \, \le\, \parallel |A|^p\parallel^{1/p} \,\, \parallel|B|^q \parallel^{1/q} $$ where $|M|:=(M^\dagger M)^\frac12$ and it holds whenever $ \parallel \cdot \parallel$ is a unitarily invariant norm. You can find a proof in the book of Bhatia Matrix Analysis.

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  • $\begingroup$ Unfortunately, the link for the proof doesn't work. $\endgroup$ – Obriareos Mar 8 '18 at 6:58
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    $\begingroup$ @Obriareos It does work on my phone and computer. In any case it's a paper by Bamugartner (2011) An Inequality for the trace of matrix products, using absolute values $\endgroup$ – lcv Mar 9 '18 at 0:16
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The closest thing I know for induced norms is the Riesz–Thorin theorem. There are other Hölder-like inequalities for matrices, for example involving Schatten norms.

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Let $A$ be a square matrix of dimension $n\times n$ and consider the following norm for $1< p<\infty$:

$$\|A\|_{p} = \max_{x \neq 0} \frac{\|Ax\|_p}{\|x\|_p}.$$ Let $\psi_p(x):=\big(|x_1|^{p-1}\operatorname{sign}(x_1),\ldots,|x_n|^{p-1}\operatorname{sign}(x_n)\big)$ and write $p'$ the Hölder conjugate of $p$. Then for every $x$ we have

$$\max_{x\neq 0 }\frac{\|Ax\|_p}{\|x\|_p} = \max_{x\neq 0 }\frac{\langle \psi_p(Ax),Ax\rangle}{\|Ax\|_p^{p-1}\|x\|_p}=\max_{x\neq 0 } \frac{\langle \psi_p(Ax),Ax\rangle}{\|\psi_p(Ax)\|_{p'}\|x\|_p}\leq \max_{x,y\neq 0} \frac{\langle y,Ax\rangle}{\|y\|_{p'}\|x\|_p} \\ \leq \max_{x,y\neq 0} \frac{\|y\|_{p'}\|Ax\|_p}{\|y\|_{p'}\|x\|_p}=\max_{x\neq 0 } \frac{\|Ax\|_p}{\|x\|_p},$$ where we used the Hölder inequality for vectors in the last inequality. It follows that $$\|A\|_p = \max_{x,y\neq 0} \frac{\langle y,Ax\rangle}{\|y\|_{p'}\|x\|_p} = \max_{x,y\neq 0} \frac{\langle A^*y,x\rangle}{\|y\|_{p'}\|x\|_p}=\|A^*\|_{p'}.$$ Now, since $\|A\|_p$ is a norm induced by a vector norm, it is submultiplicative and the spectral radius $\rho(A)$ of $A$ satisfy $\rho(A)\leq \|A\|_p$. So, we get $$\|AB\|_2^2 = \rho\big((AB)^*(AB)\big) \leq \|(AB)^*(AB)\|_p \\ \leq \|B^*A^*\|_p\|AB\|_p\leq \|B^*\|_p\|B\|_p\|A^*\|_p\|A\|_p =\big(\|A\|_p\|A\|_{p'}\big)\big(\|B\|_p\|B\|_{p'}\big).$$

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