Counter-example of a non-selfadjoint operator for which $ \left\| T \right\|= \sup_{x\in \mathcal{H},\left\| x \right\|=1} |(Tx,x)|$ does not hold.

Question

Let $\mathcal{H}$ be a Hilbert space and $T \in \mathcal{L(H )}$, if $T$ is selfadjoint, then we have : $$ \left\| T \right\|= \sup_{x\in \mathcal{H},\left\| x \right\|=1} |(Tx,x)|$$ I want a counter-example of a non-selfadjoint operator for which the equality given above does not hold.

Answer 1

Let $T$ be given by

$$T=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}.$$

Then $T$ is a non- selfadjoint operator on the real Hilbert space $\mathbb R^2$ (with the usual inner product).

We have $(Tx,x)=0$ for all $ x \in \mathbb R^2.$

Answer 2

The canonical example is $$A=\begin{bmatrix}0&1\\0&0\end{bmatrix}\in M_2 (\mathbb C). $$ You have $\|A\|=1$, while $|\langle Ax,x\rangle|\leq\frac12\,\|x\|^2$ for all $x\in\mathbb C^2 $.