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?

  • 4
    $\begingroup$ 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
    Commented Feb 27, 2014 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
    Commented Feb 27, 2014 at 21:57
  • 3
    $\begingroup$ @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
    Commented Feb 28, 2014 at 2:15
  • 1
    $\begingroup$ @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
    Commented Feb 28, 2014 at 2:36
  • $\begingroup$ related: mathoverflow.net/q/78330/84108 $\endgroup$
    – glS
    Commented Aug 1, 2023 at 19:12

3 Answers 3


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 in Bernhard Baumgartner, An Inequality for the trace of matrix products, using absolute values.

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.

  • $\begingroup$ Unfortunately, the link for the proof doesn't work. $\endgroup$
    – Obriareos
    Commented Mar 8, 2018 at 6:58
  • 1
    $\begingroup$ @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
    Commented Mar 9, 2018 at 0:16

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.


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 $$\|AB\|_2^2 = \rho\big((AB)^*(AB)\big) \leq \|(AB)^*(AB)\|_p \\ \leq \|B^*A^*\|_p\|AB\|_p\leq \|B^*\|_p\|B\|_p\|A^*\|_p\|A\|_p =\big(\|A\|_p\|A\|_{p'}\big)\big(\|B\|_p\|B\|_{p'}\big).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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