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The setup is as in this question:
Given a norm $N$ over ${\bf M}_n(\mathbb C)$, it is a natural question to find the best constant $C_N$ such that $$N([A,B])\le C_N N(A)N(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$

Equivalently, $C_N$ is the maximum of $N(AB-BA)$ provided that $N(A)=N(B)=1$.

Given examples of $C_N$ are

  • $C_N=\sqrt{2}$ if $N$ is the Frobenius norm
  • $C_N=2$ if $N$ is the operator norm $\| \cdot\|_2$
  • $C_N=4$ if $N$ is the numerical radius $r(A)=\sup\limits_{x\ne0}\dfrac{|x^*Ax|}{\|x\|^2}$ (See this answer to an MO question).

if $N$ is the induced $p$-norm, defined for $1\le p\le\infty$ by $\|A\| _p = \sup \limits _{x \ne 0} \frac{\| A x\| _p}{\|x\|_p}$, we have $C_N=2$ for $p=\infty$ (with $\|A\|_\infty $ being just the maximum absolute row sum of the matrix). Indeed, the lower bound $2$ for $\|\cdot\|_\infty $ is obtained by taking e.g. $A=\begin{pmatrix} 1&0\\1&0\end{pmatrix}$ and $B=\begin{pmatrix} 0&1\\0&-1\end{pmatrix}$, and it should be easy to prove that $2$ is also the general upper bound for $\|\cdot\|_\infty $.
Similarly, $C_N=2$ for $p=1$ (with $\|A\|_1 $ being the maximum absolute column sum of the matrix).

Knowing that $C_N\equiv2$ for $p=1,2,\infty$, is it true that the same holds for the induced $p$-norms for all $p\ge1$?

If $N$ runs over all possible matrix norms, what is the range of $C_N$? In particular, is it bounded below and/or above?

(To avoid trivialities, let's keep it homogeneous by only considering "normalized" norms, i.e. require $N(I_n)=1$. This does not seem to be part of the standard definition of a norm.)

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  • $\begingroup$ Could you give quantifiers on $n$? $\endgroup$
    – YCor
    Commented Dec 27, 2016 at 12:11
  • $\begingroup$ @YCor It seems to me that generally the $C_N$'s are independent of $n$. For a given norm, any lower bound for $n$ is also a lower bound for $n+1$, assuming that the norm behaves "reasonably" satisfying $N\begin{pmatrix} 1&\bf0^T \\ \bf0&A\end{pmatrix}=N(A)$. Well, not all norms do, now I don't know to what extent a norm can be "pathological". $\endgroup$
    – Wolfgang
    Commented Dec 27, 2016 at 13:13
  • $\begingroup$ Let $\Vert\cdot\Vert$ be a norm on a vector space, and use the same notation for the induced operator norm $\Vert A\Vert =\sup_x \frac{\Vert Ax\Vert}{\Vert x\Vert}$. Then $\Vert AB\Vert \leq \Vert A\Vert \Vert B\Vert$ so that $\Vert [A,B]\Vert\leq 2\Vert A\Vert \Vert B\Vert$ for all $A,B$. $\endgroup$
    – Lior Silberman
    Commented Dec 27, 2016 at 13:55
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    $\begingroup$ I'm not sure what you mean by "matrix norm", but if you have arbitrary norms (just using the linear structure of the space of matrices, and even with the mild normalization $N(1)=1$), it's clear that $C_N$, for each fixed $n\ge 2$ can achieve all values. Indeed for $n=2$, define $N_t(A)=\max(|a|,t|b|,t|c|,t|d|)$ where $A=\begin{pmatrix}a+b & c\\ d & a-b\end{pmatrix}$. Then $C_{N_t}$ tends to $+\infty$ when $t\in 0$ and tends to 0 when $t\to\infty$. On the other hand under the conditions $N(1)=1$, $N(AB)\le N(A)N(B)$ (e.g. operator norms), clearly $C_N\in [1,2]$. $\endgroup$
    – YCor
    Commented Dec 27, 2016 at 18:43

2 Answers 2

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A somewhat more general setting, namely, finding the best constant $C_{p,q,r}$ in \begin{equation*} \|AB-BA\|_p \le C_{p,q,r}\|A\|_q\|B\|_r, \end{equation*} for Schatten $p$,$q$,$r$-norms, is studied in this paper.

EDIT (1st Mar'17). See also this recent paper (LAA, 521(15), May 2017, Pages 263–282) that provides sharp bounds on the constant $C_{p,q,r}$ above for real matrices.

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    $\begingroup$ Thank you, that paper looks great! And some quite unexpected results :) $\endgroup$
    – Wolfgang
    Commented Jan 6, 2017 at 8:52
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Concerning your first question, as I noted in my comment above for any operator norm we have $\Vert [A,B]\Vert \leq 2\Vert A\Vert\Vert B\Vert$.

Conversely, let $A = \pmatrix{1 & 0 \\ 0 & -1}$, $B = \pmatrix{0 & 1\\1 & 0}$ so that $[A,B] = \pmatrix{0 & -2\\2 & 0} = 2\pmatrix{0 & 1\\-1 & 0}$.

Then for all $1\leq p \leq \infty$ $\Vert A\Vert_p = \Vert B\Vert_p = 1$, and $\Vert [A,B]\Vert_p = 2$

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