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

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.

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 $$||T|| \le 2N$$.
• By $T(x,y)$ do you mean $\langle Tx, y \rangle$? – Viktor Glombik Oct 8 '19 at 20:05
• Also, $w(T)$ is a set, only $r(T)$ is a number (I edited the question yesterday to reflect this). Also why does the RHS look the way it does after "taking norms"? – Viktor Glombik Oct 8 '19 at 20:12
• @BruceEvans By $T(x,y)$ do you mean $<Tx,y>$? and for polarization identity we need operator $B$ to be symmetric bilinear form, can you please help me to see why $B$ is symmetric bilinear? – Aarna Feb 23 at 21:14