Let $H$ be a Hilbert space and $A \in B(H)$. How to prove that if $|\lambda| = \|A\|$ and $\lambda \in W(A)$, then $\lambda \in \sigma_p(A)$?
Here $W(A)$ is the numerical range of the operator $A$: $$W(A):= \big\{(Ax,x): \|x\| = 1\big\}$$ and $\sigma_p(A)$ is the point spectrum of $A$.
I have no idea how to prove this. Can you help me with this?
There is $x$ with $\|x\|=1$ such that $$ \lambda = \langle Ax,x\rangle. $$ By the Cauchy-Schwarz inequality, $$ |\langle Ax,x\rangle| \le \|A\|= |\lambda| $$ with equality if and only if $Ax$ and $x$ are linearly dependent. This implies that $Ax$ and $x$ are linearly dependent: $c_1 Ax = c_2 x$. Since $x\ne0$, it follows $c_1\ne0$, and $x$ is an eigenvector to the eigenvalue $c_2/c_1$. Multiplying the equation with $x$ implies $c_1 \lambda = c_2$, and $\lambda$ is an eigenvalue.