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!


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 $||T|| \le 2N$.

  • $\begingroup$ My proof works for real and complex scalars. Complex scalars mainly allow more to be done, starting with the polarization identity for a non self-adjoint (non-symmetric) operator not even existing with real scalars. $\endgroup$ Sep 20 '16 at 13:21
  • $\begingroup$ By $T(x,y)$ do you mean $\langle Tx, y \rangle$? $\endgroup$
    – Ramanujan
    Oct 8 '19 at 20:05
  • $\begingroup$ 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"? $\endgroup$
    – Ramanujan
    Oct 8 '19 at 20:12
  • $\begingroup$ @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? $\endgroup$
    – Aarna
    Feb 23 '20 at 21:14

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