# How to prove this for the spectrum of a self-adjoint operator?

Let $$T$$ be a bounded self-adjoint operator. Prove:

A number $$\lambda \in \mathbb{R}$$ belongs to the spectrum of $$T$$ if and only if $$\mathbb{1}_{(\lambda - \epsilon, \lambda + \epsilon)} (T)$$ is non-zero for any $$\epsilon > 0$$.

I think I should approximate the characteristic function of the interval with a continuous function. But I'm still not sure how to prove this. Help is appreciated.

If $$\lambda$$ belongs to the spectrum $$\sigma(T)$$ of $$T$$, then $$(\lambda-\varepsilon, \lambda+\varepsilon)$$ is a non-empty open subset of $$\sigma(T)$$. By Urysohn's lemma, there exists a non-zero continuous function $$f$$ such that $$0\leq f\leq 1_{(\lambda-\varepsilon, \lambda+\varepsilon)}$$. Therefore, $$1_{(\lambda-\varepsilon, \lambda+\varepsilon)}(T)\geq f(T)$$. Since functional calculus of continuous functions is an injective *-homomorphism, $$f(T)\geq 0$$ and $$f(T)\neq 0$$. Thus $$1_{(\lambda-\varepsilon, \lambda+\varepsilon)}(T)\neq 0$$.
If $$\lambda$$ does not belong to the spectrum $$\sigma(T)$$ of $$T$$, one can find a $$\varepsilon_0>0$$ such that $$(\lambda-\varepsilon_0, \lambda+\varepsilon_0)\cap \sigma(T)=\emptyset$$. So $$1_{(\lambda-\varepsilon_0, \lambda+\varepsilon_0)}=0$$.
• Why is $(\lambda-\varepsilon,\lambda+\varepsilon)$ a subset of $\sigma(T)$? – Jarne Renders Nov 21 at 20:06
• Precisely, I should say that $(\lambda-\varepsilon, \lambda+\varepsilon)\cap \sigma(T)$ is a non-empty relatively open subset of $\sigma(T)$. – C.Ding Nov 21 at 20:30