We denote the numerical range of a complex square matrix $A \in \mathbb{C}^{n\times n}$ by $W(A)$.

Let $A \in \mathbb{C}^{n\times n}$ and let $f: \mathbb{C} \to \mathbb{C}$ be, say, an entire function. It is easy to see that a mapping theorem for the numerical range in the sense of $W(f(A)) = f(W(A))$ does not hold, in general. The following example gives a bit more insight:

Example. Let $A$ be the $3\times 3$-diagonal matrix with diagonal entries $0, \pi i, 2\pi i$ and let $f(z) = e^z$. Then $W(f(A))$ is the line segment $[-1,1]$, but $f(W(A))$ is the complex unit circle. In fact, this even shows that...

  • we don't have $W(f(A)) = \operatorname{conv}(f(W(A)))$, in general (where $\operatorname{conv}$ denotes the convex hull).

  • we don't have an inclusion theorem of the kind $W(f(A)) \subseteq f(W(A))$, in general.

Now, it seems natural to ask:

Question. Is there also a counterexample known for the more general inclusion \begin{align*} W(f(A)) \subseteq \operatorname{conv}(f(W(A))) \qquad (*) \end{align*} or is it an open problem whether $(*)$ holds?

Note. It is certainly not known that $(*)$ is true, due to the following reasoning:

(i) If $(*)$ is true, this immediately implies $w(f(A)) \le \sup_{z \in W(A)} \lvert f(z) \vert$, where $w$ denotes the numerical radius.

(ii) By the well-known inequality $\|B\| \le 2w(B)$ for every matrix $B$ (where $\|\,\cdot\,\|$ denotes the norm induced by the $2$-norm on $\mathbb{C}^n$), $(*)$ would thus imply that Crouzeix's conjecture \begin{align*} \|f(A)\| \le 2 \sup_{z \in W(A)} \lvert f(z) \vert \end{align*} is true.

Remark. It is easy to see that $(*)$ holds for every normal matrix (which also explains why it is true in the above example), but I could not even figure out whether it holds for $2 \times 2$ Jordan blocks.

Disclaimer. I am, of course, not asking whether Crouzeix's conjecture is true. I am asking whether the more general assertion $(*)$ is known to be false or whether it is an open problem.

  • 2
    $\begingroup$ You can try $N=[\begin{smallmatrix} 0&1\\0&0\end{smallmatrix}]$ and $f(z)=z+z^2+\cdots+z^m$. On the one hand $f(N)=N$ and $W(N)=B(0,1/2)$. On the other hand $f(z)\approx z(1-z)^{-1}=(1-z)^{-1} -1$ and ${\rm co}f(B(0,1/2))$ misses $-1/2$. $\endgroup$ – Narutaka OZAWA Mar 23 '19 at 12:38
  • $\begingroup$ @NarutakaOZAWA: Great example, thank you! If you add it as an answer I will, of course, accept it. $\endgroup$ – Jochen Glueck Mar 24 '19 at 17:02

Let $N=[\begin{smallmatrix} 0&1 \\ 0&0 \end{smallmatrix}]$ and $f(z)=z+z^2+\cdots+z^m$. On the one hand, $f(N)=N$ and $W(f(N))=W(N)=B(0,1/2)$, the closed ball of center $0$ and radius $1/2$. On the other hand, since $f(z) \approx \sum_{k\geq1} z^k = z(1-z)^{-1}=(1-z)^{-1}-1$ to within $2^{-m}$ for $z \in B(0,1/2)$ and $\inf_{z\in B(0,1/2)}\Re (1-z)^{-1}-1 = -1/3$, the convex hull of $f(B(0,1/2))$ misses $-1/2$, as long as $m\geq3$.

P.S. Thank you for informing me of Crouzeix's conjecture. It looks interesting.

| cite | improve this answer | |

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.