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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.

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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 $1_{(\lambda-\varepsilon_0, \lambda+\varepsilon_0)}=0$.

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  • $\begingroup$ Why is $(\lambda-\varepsilon,\lambda+\varepsilon)$ a subset of $\sigma(T)$? $\endgroup$ – Jarne Renders Nov 21 at 20:06
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    $\begingroup$ Precisely, I should say that $(\lambda-\varepsilon, \lambda+\varepsilon)\cap \sigma(T)$ is a non-empty relatively open subset of $ \sigma(T)$. $\endgroup$ – C.Ding Nov 21 at 20:30

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