Let $A$ be a bounded operator on hilbert space $H$. If $\lVert A\rVert>1$, there $\exists$unit vectors $x,y$ such that $\langle Ax,y\rangle<-1$

Question

I know that for $A\in\mathcal{B}(H)$ we have $\lVert A\rVert=\text{sup}\{|\langle Ax,y\rangle|:\ \lVert r\rVert=\lVert y\rVert=1\}$

If $\lVert A\rVert>1$, there there exist unit vectors $x,y\in H$ such that $|\langle Ax,y\rangle|>1$. But this $\langle Ax,y\rangle$ may not be real, if it is then either $\langle Ax,y\rangle <-1$ or $\langle Ax,-y\rangle <-1$. So I guess that we need to adjust $x,y$ such that it becomes real.

Can anyone suggest me a wayout to prove the statement? Thanks for your help and valuable suggestions in advance.

Answer 1

Let $\langle Ax, y \rangle=re^{i\theta}$ with $r>1$ and $\theta$ real. Let $u=-e^{i\theta} x$. Then, $\langle Au, y \rangle=-r <-1$ and $\|u\|=1$.