If $H$ is a Hilbert space and $T : H \to H$ is a self-adjoint bounded operator, under what conditions can we guarantee that $\sup\limits_{||x|| = 1} \langle Tx, x \rangle \le ||T|| < \infty$ is attained by some $x$?
It's enough that $T$ is compact, but I don't have this condition. I'm looking at $H = L^2([0,1])$ and an operator which isn't compact.
Let $M=\sup_{\|x\|=1}\langle Tx,x\rangle.$ The value $M$ is attained if and only if $M$ is an eigenvalue of $T.$ Indeed assume $$\langle Tx,x\rangle =M,\quad \|x\|=1 $$ The operator $A=MI-T$ is positive and $$\langle Ax,x\rangle =0$$ Hence $Ax=0,$ i.e. $Tx=Mx.$