# If $A$ is self-adjoint, then $\left\|A\right\|=\max_{\lambda\in\sigma(A)}|\lambda|$

Let $$H$$ be a $$\mathbb R$$-Hilbert space, $$A\in\mathfrak L(H)$$ be self-adjoint with $$\left\|A\right\|_{\mathfrak L(H)}\le1$$, $$\sigma(A)$$ and $$r(A)$$ denote the spectrum and spectral radius of $$A$$, respectively, $$\lambda_{\text{min}}:=\inf_{\lambda\in\sigma(A)}\lambda$$, $$\lambda_{\text{max}}:=\sup_{\lambda\in\sigma(A)}\lambda$$, $$\operatorname{abs gap}A:=1-\sup_{\lambda\in\sigma(A)}|\lambda|$$

I want to show that $$1-\left\|A\right\|_{\mathfrak L(H)}=\operatorname{abs gap}A.\tag1$$

We know that $$\left\|A\right\|_{\mathfrak L(H)}=r(A)=\sup_{\left\|x\right\|_H\le1}\left|\langle Ax,x\rangle_H\right|\tag2$$ and $$\sup_{\lambda\in\sigma(A)}|\lambda|\le r(A)\tag3.$$ Maybe we can show that $$1-\operatorname{abs gap}A=\max\left(\left|\lambda_{\text{min}}\right||,\lambda_{\text{max}}\right).\tag4$$

How can we finish up and are $$\inf_{\left\|x\right\|_H\le1}\langle Ax,x\rangle_H$$ and $$\sup_{\left\|x\right\|_H\le1}\langle Ax,x\rangle_H$$ relaed to $$\lambda_{\text{min}}$$ and \lambda_{\text{max}}?

I know the identity $$\left\|A\right\|_{\mathfrak L(H)}=\sup_{\lambda\in\sigma(A)}|\lambda|$$, but I'm not sure whether we need that $$H$$ is a complex Hilbert space in order for this identity to hold.