Let $A$ be a bounded self-adjoint operator on a Hilbert space $H$. Suppose that there exist $\lambda,c\in\mathbb{R}$ and $u\neq 0$ in $H$ such that
$$\left\|(A-\lambda)u\right\|_H\leq c\|u\|_H.$$
Question: Does there exist $\lambda'$ in the spectrum of $A$ such that $|\lambda-\lambda'|\leq c$?
Thoughts: If $A$ is a diagonal matrix I can see why this is true, but I'd like to prove it for general $A$.
Let $\sigma(A)$ denote the spectrum of $A$. Suppose to the contrary that $\left|\lambda-\lambda'\right|>c$ for all $\lambda'\in\sigma(A)$.
Then the contradiction follows from the spectral theorem, by which we have
$$\left\|(A-\lambda)u\right\|_H^2=\int_{\sigma(A)} (\lambda-\lambda')^2\,d\mu_u(\lambda’)>c^2\int_{\sigma(A)}d\mu_u(\lambda’)=c^2\left\|u\right\|_H^2.$$
Here $d\mu_u$ is the unique positive measure associated to $A$ and $u$ satisfying the property that for all $\varphi\in C_c(\mathbb{R})$, $$\langle\varphi(A)u,u\rangle_H,$$ $\varphi(A)$ being defined by the continuous functional calculus.