29

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.

the_fox
  • 5,928
Hui Yu
  • 15,469
  • 3
    "we use our procedure to 'explain' the convexity of the numerical range (and some of its generalizations) of a complex matrix" -- from Automatic Convexity by Akemann and Weaver. Also, journal link –  Jul 17 '12 at 00:39
  • Please use \langle and \rangle for angle brackets, not < and > (which give inequality symbols). – Zev Chonoles Jul 17 '12 at 01:05
  • 5
    Not a conceptual explanation as you seem to be expecting, but still a very nice argument is given in Gustafson, The Toeplitz-Hausdorff theorem, Proc. Amer. Math. Soc. 25 (1970), 203-204. For those who read German: Toeplitz explains in section 6 at the end of his original paper Das algebraische Analogon zu einem Satze von Fejér, Math. Z. 2 (1918), no. 1-2, 187–197, how he found the result by translating work of Fejér on Fourier series into a geometric setting. – t.b. Jul 17 '12 at 09:04
  • @t.b. Although much shorter than most of the proofs I have seen, the first paper is still a computational one. Unfortunately I cannot read German, could you please be so kind to provide a little bit of the taste of what is Fejer's work that Toeplitz is referring to? – Hui Yu Jul 17 '12 at 14:18
  • 1
    Fejér's theorem that inspired Toeplitz is: Let $f$ and $g$ be two $2\pi$-periodic continuous and real-valued functions. Consider the convex hull $C$ of the curve $t \mapsto (f(t),g(t))$ in the real plane. Then for each $n$ the curve $t \mapsto (\sigma_n(f,t), \sigma_n(g,t))$ given by the $n$th Fejér sums of $f$ and $g$ (the arithmetic means of the first $n+1$ Fourier approximations of $f$ and $g$) lies entirely in $C$. Toeplitz recast this result first in terms of Laurent series (his thesis) and then in terms of infinite binary forms. – t.b. Jul 17 '12 at 15:08
  • 2
    My personal relationship with this fact is a bit dialectic: after the Thesis is presented I find it weird that something homogeneous of degree two, which I feel should thus behave like a polynomial, could ever produce a convex set! But then I think of the case of a normal matrix, where the squares conspire to produce a convex combination of the eigenvalues, and then I feel a lot better! – Ruy Nov 17 '20 at 00:58
  • 1
    This post is relevant – Ben Grossmann Dec 31 '21 at 16:29

1 Answers1

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.

Harish Chandra Rajpoot
  • 38,565
Bikhu
  • 88