Why is the numerical range of an operator convex?

Question

Let $T$ be a Hilbert space operator. Its numerical range is \begin{equation} W(T)=\{\langle Tx,x\rangle:\|x\|=1\}.\end{equation}

It is a well-known fact that $W(T)$ is a convex subset of the complex plane. However, every proof I know is by brute force computation. First for $2\times 2$ matrices, then the general case.

Even though the computation can be carried out in clever ways, it still fails to provide some explanation why this is true. What is the link between this result and other concepts of the theory?

I wonder whether there is any conceptual explanation for this result. I do not ask the explanation to be rigorous, just some ideas.

Answer 1

Let $\mathcal{H}$ be a complex separable Hilbert space and $T$ be a bounded linear operator on $\mathcal{H}$. Let us assume in advance that, for a $2\times 2$ complex matrix $A$ the $W(A)$ is an elliptic disc (The result is true but I am unable to show here because it will make the proof much bigger).

Now we have to show that for any $\alpha,\beta \in W(T)\implies\lambda \alpha+(1-\lambda)\beta\in W(T)$, where $\lambda\in (0,1)$. Let $\alpha=\langle Tf,f\rangle$ and $\beta=\langle Tg,g\rangle$ where $\|f\|=\|g\|=1$. Clearly $f,g$ are linearly independent unit vectors otherwise, $\alpha=\beta$. Denote $V=\mbox{span}\{f,g\}$ and $\dim V=2$. Since $V$ is closed, therefore there exists an orthogonal projection $P_V$ onto $V$.

Now the compression $A=P_V T|_V$ of $T$ is a two dimensional operator. So, $W(A)$ is an elliptic disc from our initial assumption. Again $\langle A f,f\rangle= \langle Tf,f\rangle=\alpha$ and $\langle A g,g\rangle= \langle Tg,g\rangle=\beta$. Therefore, $\lambda \alpha+(1-\lambda)\beta\in W(A)$, where $\lambda\in (0,1)$. Also we know that $W(A)\subseteq W(T)$. Thus, $\lambda \alpha+(1-\lambda)\beta\in W(T)$. Hence we get our required result.

Answer 2

I am not sure if this counts as a conceptual explanation or not, but Yik-Hoi Au-Yeung has given a simple proof based on his diametric exclusion principle. See Y.-H. Au-Yeung, A simple proof of the convexity of the field of values defined by two Hermitian forms, Aequationes mathematicae 12: 82-83 (1975) (available from the European Digital Mathematics Library).

Au-Yeung’s diametric exclusion principle is set in a left vector space $V$ of possibly infinite dimension over $F=\mathbb R,\mathbb C$ or $\mathbb H$. In essence and to water things down a bit, it simply states that if $n\ge3$ and $A,B\in M_n(\mathbb R)$ are two symmetric matrices such that $(x^\top Ax,\,x^\top Bx)\ne(0,0)$ for every nonzero vector $x$, and if we define a function $f:S^{n-1}\to S^1$ $$ f(x)=\frac{(x^\top Ax,\,x^\top Bx)}{\sqrt{(x^\top Ax)^2+(x^\top Bx)^2}} $$ between the unit sphere in $\mathbb R^n$ and the unit circle in $\mathbb R^2$, then there does not exist any pair of unit vectors $u$ and $v$ such that $f(u)=-f(v)$. That is, no two points in the image of $f$ lie on the opposite sides of the same diameter.

If $n\ge2$ and we view $\mathbb C^n$ as a real vector space of dimension $2n\ge4$, then Au-Yeung’s diametric exclusion principle implies that for any pair of Hermitian matrices $H_1,H_2\in M_n(\mathbb C)$ such that $g(z):=(z^\ast H_1z,\,z^\ast H_2z)\ne(0,0)$ for every nonzero vector $z$, there does not exist any positive real number $\varepsilon$ and any pair of nonzero vectors $u$ and $v$ such that $g(u)=-\varepsilon g(v)$.

You have mentioned that every proof you know for the convexity of $W(T)$ is first done for $2\times2$ matrices. Au-Yeung’s convexity proof does not assume that $T$ is defined on a finite-dimensional Hilbert space, but it really doesn’t hurt to consider only the two-dimensional case, because $[\langle Tu,u\rangle,\langle Tv,v\rangle]\subseteq W(T)$ if the numerical range of the complex $2\times2$ matrix $$ \pmatrix{\langle u,Tu\rangle& \langle u,Tv\rangle\\ \langle v,Tu\rangle&\langle v,Tv\rangle\rangle} $$ is convex. So, the problem is inherently two-dimensional and for simplicity we may assume that $T\in M_2(\mathbb C)$. Let $T=H_1+iH_2$, where $H_1$ and $H_2$ are Hermitian. Pick any two points $u^\ast Tu=a_1+ib_1$ and $v^\ast Tv=a_2+ib_2$ in $W(T)$ (where $u$ and $v$ are unit vectors). Let $a+ib$ be their mid-point. Let also $\widehat{H_1}=H_1-a$, $\widehat{H_2}=H_2-b$. If $a+ib\not\in W(T)$, we will have $(z^\ast \widehat{H_1}z,\,z^\ast \widehat{H_1}z)\ne(0,0)$ for every nonzero vector $z$ but $$ (u^\ast \widehat{H_1}u,\,u^\ast \widehat{H_2}u) =(a_1-a,b_1-b) =-(a_2-a,b_2-b) =-(v^\ast H_1v,\,v^\ast H_2v), $$ which violates Au-Yeung’s diametric exclusion principle. Hence the mid-point of any two points in $W(T)$ is again a point in $W(T)$. By a continuity argument, $W(T)$ is convex.

Remark. Au-Yeung first proved and used his diametric exclusion principle to create a neat and simple proof of Finsler’s lemma. See Y.-H. Au-Yeung, A theorem on a mapping from a sphere to the circle and the simultaneous diagonalization of two Hermitian matrices, Proc. Amer. Math. Soc., 20: 545-548 (1969). While the exclusion principle is correct, his proof for it is not. In particular, he builds his proof on his lemma 1, which states that

For any real numbers $a_0,\ldots,a_4,b_0,\ldots,b_4$ the following system of equations \begin{align*} a_0(x^2-y^2)+a_1xy+a_2xz+a_3yz+a_4z^2&=0,\\ b_0(x^2-y^2)+b_1xy+b_2xz+b_3yz+b_4z^2&=0,\\ \end{align*} has nontrivial real solutions.

But this lemma is obviously false, as shown by the classical counterexample $x^2-y^2=0=xy$ to the two-dimensional case of (a formulation of) Finsler’s lemma.