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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.

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    $\begingroup$ 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 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 at 17:02
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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.

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