# If $A$ is self-adjoint, then $\left\|A\right\|=\sup_{x\in H\setminus\{0\}}\frac{\langle Ax,x\rangle}{\left\|x\right\|^2}$

Let $$H$$ be a $$\mathbb R$$-Hilbert space and $$A\in\mathfrak L(H)$$ be self-adjoint. I want to show that $$\left\|A\right\|_{\mathfrak L(H)}=\sup_{x\in H\setminus\{0\}}\frac{\langle Ax,x\rangle_H}{\left\|x\right\|_H^2}.\tag1$$ How can we do that? I know that for any norma operator on a Hilbert space, the oprator norm is equal to the spectral radius. However, that doesn't seem to help.

EDIT: I guess we need to assume that $$A$$ is nonnegative (i.e. $$\langle Ax,x\rangle_H\ge0$$ for all $$x\in H$$.

• What is the definition of the operator norm you are using? Nov 11, 2019 at 23:15
• @DanielC $\left\|A\right\|_{\mathfrak L(H)}=\sup_{x\in H\setminus\{0\}}\frac{\left\|Ax\right\|_H}{\left\|x\right\|_H}.$ Nov 12, 2019 at 5:14
• Notice that a non-negative bounded linear operator is self-adjoint (Lemma 13.12). Mar 1, 2020 at 23:44
• Why isn't the answer accepted? Mar 1, 2020 at 23:45

Let $$B\in\mathfrak L(H)$$. By the Cauchy-Schwarz inequality, $$c:=\sup_{x\in H\setminus\{0\}}\frac{|\langle Bx,x\rangle_H|}{\left\|x\right\|_H^2}\le\left\|B\right\|_{\mathfrak L(H)}=\sup_{\left\|x\right\|_H,\:\left\|y\right\|_H\:\le\:1}|\langle Bx,y\rangle_H|.\tag2$$ Now asssume that $$B$$ is self-adjoint. Then, $$\langle B(x+y),(x+y)\rangle_H-\langle B(x-y),x-y\rangle_H=4\langle Bx,y\rangle_H\tag3$$ and hence $$|\langle Bx,y\rangle_H|\le c\frac{\left\|x+y\right\|_H^2+\left\|x-y\right\|_H^2}4=c\frac{\left\|x\right\|_H^2+\left\|y\right\|_H^2}2\tag4$$ by the parallelogram law for all $$x,y\in H$$, i.e. $$\sup_{\left\|x\right\|_H,\:\left\|y\right\|_H\:\le\:1}|\langle Bx,y\rangle_H|\le c.\tag5$$ Thus, $$\left\|B\right\|_{\mathfrak L(H)}=c.\tag6$$