bilenear form extrema equality

Question

Let $A$ a self-adjoint positive contnious operator from $H$ into $H$. Do we have: $$sup(Ax,x) = \sup (Ax,y)$$ for all $||x|| = 1,||y|| = 1$?. Thanks.

Answer 1

Let $M=\sup\{<Ah,h>,||h||=1\}$. If $||h||=1$, then $<Ah,h>\leq||A||$, so $M\leq||A||$.

And $\forall f\in H, <Af,f>\leq M$. If $||h||=||g||=1$, then : $$<A(h\pm g),h\pm g>=<Ah,h>\pm<Ah,g>\pm <g,A^*h>+<Ag,g>$$ But $A^*=A$, so : $$<A(h\pm g),h\pm g>=<Ah,h>\pm2\Re<Ah,g>+<Ag,g>$$ By substracting these inequalities for $+$ and $-$ we get : $$4\Re<Ah,g>=<A(h+ g),h+ g>-<A(h- g),h- g>\\\leq M(||h+g||^2+||h-g||^2)$$ We parallélogram law we have : $$4\Re<Ah,g>\leq2M(||h||^2+||g||^2)=4M$$

Let $\theta \in [0,2\pi]$ such that $<Ah,g>=e^{i\theta}|<Ah,g>|$. We apply the previous inequlity with $e^{i\theta}h$ instead of $h$ and we get that $|<Ah,g>|\leq M$ when $||h||=||g||=1$. We take the supremum on $g$ and $h$, and we get $||A||\leq M$.

Finally $||A||=M$.