# Bounded spectrum implies bounded operator?

Let $T$ be a self-adjoint operator with bounded spectrum $\sigma(T)$.

Does this imply that $T$ is bounded?

I would say: Yes!

My attempt: $$||T|| = sup_{||x||=1} \langle Tx,x \rangle =sup_{||x||=1} \int_{\sigma(T)} \lambda d\mu_{x,x}(\lambda) \le \sup_{t \in \sigma(T)}|t| \quad sup_{||x||=1} ||\mu_{x,x}|| \le \sup_{t \in \sigma(T)}|t| < \infty .$$

I am new to the spectral theorem, so I don't know if my proof is correct.

Almost right. The trouble is that this doesn't allow for negative $\lambda$'s. You could use

$$\|T\|^2 = \sup_{\|x|=1} \langle Tx, Tx \rangle = \sup_{\|x\|=1} \int_{\sigma(T)} \lambda^2 \; d\mu_{x,x}(\lambda) \ldots$$