Toeplitz–Hausdorff theorem

Remarks

Proof idea: take two points $\ip{Tx}{x}$ and $\ip{Ty}{y}$ of $W(T)$ and let $M$ be the span of the unit vectors $x$ and $y$. The numerical range of the compression $P_M T|_M$ of $T$ to $M$ is contained in $W(T)$ and contains both points, so it is enough to treat a $2 \times 2$ matrix. For a $2 \times 2$ matrix, a direct computation shows that the numerical range is a closed elliptical disc with foci at the two eigenvalues. The disc may degenerate to a segment or a point.

Example: for $T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ on $\C^2$ and a unit vector $x = (x_1, x_2)$, one gets $\ip{Tx}{x} = \abs{x_1}^2 - \abs{x_2}^2$, so $W(T) = [-1, 1]$, a convex set. The set $\set{1, -1}$ of eigenvalues is not convex, but $W(T)$ fills in the segment between them.

Going deeper

For a student

The numerical range contains the eigenvalues of $T$, and its closure contains the spectrum. With convexity, the closure of $W(T)$ contains the convex hull of the spectrum, with equality for normal operators. The numerical radius $w(T) = \sup \set{\abs{z} : z \in W(T)}$ satisfies $\tfrac12 \norm{T} \le w(T) \le \norm{T}$, so it is a norm on bounded operators equivalent to the operator norm.