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.
