spectrum of two operators on Banach space

Question

Question: Let $E$ be a Banach space, $T\in \mathcal{B}(E)$. Prove that: for all $\epsilon > 0$, there exist $\delta > 0$, such that for all $S\in \mathcal{B}(E)$, if$\|T-S\| < \delta$, then $$\sigma(S)\subset \lbrace \lambda \in \mathbb{K}: d(\lambda, \sigma(T))< \epsilon \rbrace$$ where $\sigma(T)$ and $\sigma(S)$ is the spectrum set of $T$ and $S$.

My attempt: Let $\lambda \in \sigma (S)$, we can suppose that $\lambda \notin \sigma(T)$. For all $\mu \in \sigma(T)$, I know that $\|S\| \geq |\lambda|$, $\|T\| \geq |\mu|$, and $$|\lambda- \mu|=\|\lambda-\mu\|=\|\lambda - S + S - T + T -\mu\|\le \|\lambda-T\|+\|S-T\|+\|T- \mu\|$$ or $$|\lambda- \mu|=\|\lambda-\mu\|=\|\lambda - S + S - T + T -\mu\|\le \|\lambda-S\|+\|S-T\|+\|S- \mu\|$$

I don't know how to use the $\|S\| \geq |\lambda|$, $\|T\| \geq |\mu|$ to compute the $\|\lambda-S\|, \|S- \mu\|$.

I had seen two related questions Spectrum in Banach Algebra and An exercise about the spectrum of an element in Banach algebra. . But i still have no idea about my question.

Thanks in advance.

Answer 1

Finally i understood the answer written by Jonas Meyer for the questionSpectrum in Banach Algebra, so i write the answer for my question as follow:

Proof: Let $U_\epsilon =\lbrace \lambda \in \mathbb{K}: d(\lambda, \sigma(T))< \epsilon \rbrace$, obviously $U_\epsilon$ is a open set. For the large enough $R > 0$, $\overline{B(0,R)} \setminus U_\epsilon$ is a bounded closed set in $\mathbb{K}$, thus is a compact set.

Let $f: \rho(T) \longrightarrow \mathbb{K}: z \mapsto f(z)=\|(T-z)^{-1}\|$. Due to properties of resolvent, we know $f$ is continuous. And $$\lim_{|z|\rightarrow \infty}f(z)=0.$$ It means that for any $z_0\in \mathbb{K}\setminus U_\epsilon$, there exists $R>0$ such that for all $|z|>R$, there is $f(z)=|f(z)|\le |f(z_0)|=f(z_0)$. Thus $f$ could take the maximum value on the compact set $\overline{B(0,R)} \setminus U_\epsilon$, thus could take the maximum value $C>0$ on the $\mathbb{K}\setminus U_\epsilon\subset \rho(T)$.

Now we let $\delta = \frac{1}{C}$, For all $\lambda \in \mathbb{K}\setminus U_\epsilon$, then $\|(T-\lambda)^{-1}\|\le C$,thus $$\|(\lambda-T)-(\lambda-S)\|= \|S-T\|<\frac{1}{C}\le \frac{1}{\|(\lambda-T)^{-1}\|}.$$ As $\lambda-S=(\lambda-T)+(T-S)=(\lambda-T)(1+(\lambda-T)^{-1}(T-S))$,but $\|(\lambda-T)^{-1}(T-S)\|\le \|(\lambda-T)^{-1}\|\|(T-S)\| <1$, thus $1+(\lambda-T)^{-1}(T-S)$ is invertible, then $\lambda-S$ is invertible, i.e. $\lambda \in \rho(S)$. Thus $\mathbb{K}\setminus U_\epsilon \subset \rho(S)$, i.e. $\sigma(S) \subset U_\epsilon$.