Counterexample to a formula for the norm of a self-adjoint operator on a Hilbert space

Question

In my functional analysis class, I have come across this exercise

For a Hilbert space $H$ and $A$ a self-adjoint operator on $H$ one has the familiar identity $$\|A\|=\sup_{\|x\|=1} \lvert \langle Ax,x \rangle \rvert $$
we are asked to show a counterexample to this when $A$ is not self-adjoint.

I cannot think of any counterexample to this. All help is appreciated.

Answer 1

On $\mathbb R^{2}$ define $A(x,y)=(-y,x)$. Then $\|A\|=1$ and $\langle Av, v \rangle=0$ for all $v$.