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

The answer is known at least in the following cases:

- the operator norm $\|A\|_2=\sup\frac{\|Ax\|_2}{\|x\|_2}$ where the norm over $\mathbb C^n$ is the standard Hermitian $\|x\|_2^2=\sum_j|x_j|^2$. Then $$\|[A,B]\|_2\le2\|A\|_2\|B\|_2$$ is optimal for $n\ge2$.
- the Frobenius norm $\|A\|^2_F=\sum_{i,j}|a_{ij}|^2$. Then a theorem by Böttcher & Wentzel (2008) tells us that $$\|[A,B]\|_F\le\sqrt2\|A\|_F\|B\|_F,$$ and again this is optimal.

I have a third norm in mind, yet of a different nature: the

numerical radius$$r(A)=\sup_{x\ne0}\frac{|x^*Ax|}{\|x\|^2}.$$ This is the smallest radius of a disk $D(0;r)$ containing thenumerical range(or Hausdorffian) of the matrix. What is the optimal constant $C_{nr}$ such that $r([A,B])\le C_{nr}r(A)r(B)$ for all $A,B$ in ${\bf M}_n(\mathbb C)$ ?

Let me point out that $r$ is not submultiplicative. We have at best $r(MN)\le 4r(M)r(N)$, which gives by the triangle inequality $r([A,B])\le8r(A)r(B)$, but this is certainly not optimal. However, it is a *super-stable* norm, in the sense that $r(M^k)\le r(M)^k$ for every $k\ge1$.

This question naturally extends to $n$-commutators, in the spirit of my previous question Standard polynomials applied to matrices .

**Edit**. See below Piotr Migdal's answer and my adaptation of it. It gives $C_{nr}=4$.