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Let $L(H)$ be the set of bounded linear operators on a Hilbertspace. From two different resources i found the definition of the resolvent set for an operator $T \in L(H)$ like the following: $$\rho_1(T):=\{\lambda \in \mathbb{C} : \exists (\lambda - T)^{-1} \in L(H)\}$$ or $$\rho_2(T):=\{\lambda \in \mathbb{C} : (\lambda-T)\ is\ bijective\ and\ (\lambda-T)^{-1}\ is\ continous \}$$

It's obvious that $\rho_2(T) \subseteq \rho_1(T)$. What i dont get is the other direction.

From the bounded inverse theorem it follows that if $(\lambda-T)$ is linear bounded bijective its inverse is bounded too. Which means the inverse is continous in this case.

So my problem is the following statement: Given an operator $T\in L(H)$ and $\lambda \in\mathbb{C}$ such that $\exists (\lambda-T)^{-1}\in L(H)$ it follows that $(\lambda-T)\in L(H)$ is bijective.

Partly proof: Linearity of $(\lambda-T)$ is clear. From the boundedness of $T$ and $\|(\lambda-T)x\| \le \|\lambda I x + (-T)x\|\le |\lambda|\|x\| + \|T\|\|x\|\le C \|x\|$ it follows that $(\lambda-T)$ is bounded too.

Surjectivity: Let $y\in H$ then it follows that there exists $x\in H$ such that $(\lambda-T)^{-1}y=x$ $\Leftrightarrow$ $y=(\lambda-T)x$.

Regarding injectivity i'm not really sure if my argumentation is valid: Let $x_1,x_2\in H$ such that $(\lambda-T)x_1=y=(\lambda-T)x_2$. Because the inverse exists we have $\{x_1,x_2\} \in (\lambda-T)^{-1}y$. Since the inverse is unique linear and bounded it follows $x_1=x_2$. $\Box$

To be honest, if my thoughts on the injectivity are correct i dont understand them correctly. What exact property is it that injectivity follows? And how to proof this in more detail? Furtherly does this mean, that given an inverse operator $S^{-1}\in L(H)$ it follows that $S\in L(H)$ must be bijective? So the existence of an inverse linear bounded operator means bijectivity for the operator itself?

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2 Answers 2

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An inverse operator $T^{-1} \in L(H)$ is an operator such that:

\begin{align*} T^{-1}T = I = TT^{-1} \end{align*}

Thus injectivity of $(\lambda -T)$ is proved by noting that:

\begin{align*} (\lambda -T)x_1 = (\lambda - T)x_2 \Rightarrow (\lambda - T)^{-1} (\lambda -T)x_1 = (\lambda - T)^{-1} (\lambda -T)x_2 \Rightarrow x_1=x_2 \end{align*}

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A possible way to go with the proof of injectivity $$(\lambda -T)x_1=(\lambda -T)x_2,$$ $$\lambda I(x_1-x_2)=T(x_1-x_2),$$ $$(\lambda - T)(x_1-x_2)=0.$$ This is equivalent to the fact that $x_1-x_2 \in \text{Ker}(\lambda I-T)$. However since the operator $(\lambda - T)$ is invertible, one can conclude that $$\text{Ker}(\lambda I-T)=0$$ and hance $x_1=x_2$. As you amy see from above, the result is thank to the linearity of the operator. Try to compare it with finite-dimensional case, i.e., matrix operators.

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