equality of norm and numerical radius for normal (or even self-adjoint) operators on Hilbert space

Question

Let $T:\mathcal{H}\to\mathcal{H}$ be a bounded linear operator on a complex Hilbert space. Its numerical range is defined by $w(T) :=\{(Tx,x): ||x||\le 1\}$ and its numerical radius by $r(T) := \sup_{\lambda \in w(T)} | \lambda |$.

In Pedersen's GTM book "Analysis NOW" it is proved in proposition 3.2.27 that the numerical radius of a normal continuous operator $T$ is equal to its usual operator norm. I have problems to follow his argument which relies on proposition 3.2.26 asserting among other things that the equality of these two norms holds already for self-adjoint operators.

Can one me give either a proof or at least some reference where to find alternative proofs that the numerical radius of a self-adjoint (or normal) bounded linear operator on Hilbert space is equal to its norm?

The equivalence of numerical radius and spectral norm addresses a similar question and asks a proof of the equivalence of the two norms, but the answer provided also relies on the identity of the two norms on normal operators.

many thanks in advance!

Answer 1

If $T$ is positive, this is a form of the Cauchy-Schwarz inequality. If $T$ is self adjoint, use the polarization identity and then similar methods.

Let $B(x, y) = T(x, y)$ and $Q(x) = B(x, x)$. The polarization identity is $$ 4 B(x, y) = Q(x+y) - Q(x-y). $$ Let $N = w(T)$. Taking norms gives $$ 4 ||B(x, y)|| \le N ||x+y||^2 + N || x-y ||^2 = 2 N (||x||^2 + ||y||^2). $$ If $Tx = 0$, then $||Tx|| \le N||x||$ is trivial. Otherwise, choose $y = cTx$ where $c = \frac{\|x\|}{\|Tx\|}$ is chosen so that $||y|| = ||x||$. This gives $$ 4||x|| \cdot ||Tx|| \le 4 N ||x||^2$$ or $||T|| \le N.$ Also, $N \le ||T||$ is trivial.

This proof works with real scalars. If T is not self-adjoint, then complex scalars are needed to get a useful polarization identity. It has 4 terms instead of 2, so the above method only gives $||T|| \le 2N$.