# Mapping inclusion theorem for the numerical range

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.

• 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 $-1/2$. – Narutaka OZAWA Mar 23 '19 at 12:38
• @NarutakaOZAWA: Great example, thank you! If you add it as an answer I will, of course, accept it. – 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 $$m\geq3$$.