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I have a linear bounded operator $T:L_2(0,1) \rightarrow L_2(0,1)$ of the form

$T=I+K,$

where $I$ is the identity operator and $K$ is a Hilbert-Schmidt integral operator and hence a compact operator.

I am wondering if the following assertion is true: Operator $T$ is invertible, that is, there exists an operator $T^{-1}:L_2(0,1) \rightarrow L_2(0,1)$ such that $T^{-1}T = TT^{-1}=I$.

My Approach: I use the following results to prove the above assertion.

The following two results are from the book An Invitation to Operator Theory-Abramovich:

Lemma 2.8: For a bounded operator $S:X \rightarrow Y$ between two Banach Spaces the following are equivalent:

i) $S$ is surjective and one-to-one,

ii) $S$ is bounded below and range of $S$ $R(S)=Y$.

Corollary 2.15: For a bounded operator $S:X \rightarrow Y$ between two Banach Spaces the following are equivalent:

i) Range $R(S)$ is closed,

ii) There exists a constant $c>0$ such that for each $y \in R(S)$ there exists some $x \in X$ such that $y=Sx$ and $\|x\| \leq c \|y\|$.

Finally, I also use the result (Fredholm Alternative) from the book by Evans that the range of identity plus compact operator $R(I+K)$ is closed.

The following is my proof:

Proof: Since $K$ is compact, range of $T=I+K$, $R(T) \subseteq L_2(0,1)$ is closed. Therefore, we obtain from Corollary 2.15(ii) that the operator $T$ is surjective, i.e., for each $y \in R(T)$, there exists some $x \in L_2(0,1)$ such that $y=Tx$. Moreover, from Corollary 2.15(ii) we have that there exists a constant $c>0$ such that

$\|x\| \leq c\|Tx\|, \quad \forall x \in L_2(0,1)$.

Therefore, $Tx=0 \Rightarrow x=0$, hence, $T$ is one-to-one.

Thus, we have shown that the operator $T$ is surjective and one-to-one. Therefore, we may apply Lemma 2.8 to conclude that $R(T)=L_2(0,1)$. Finally, using these facts, we may use the open mapping theorem we conclude that the inverse $T^{-1}$ exists. End of Proof

Please critique the proof. Moreover, if there is a reference where this result is stated, I would be very grateful. Thank you.

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    I see two mistakes. First, Cor. 2.15 works for $y \in R(T)$, not for all $y \in \mathbb{L}^ 2$. Second, the inequality $\|x\| \leq \|Tx\|$ is valid only for the $x$ you choose as preimages of $y$. If the transformation is injective, then there is only one such $x$, and your proof is fine; but that's not necessarily the case. – 2017-02-17
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    So, in short: you assume surjectivity to prove surjectivity, and you assume injectivity to prove injectivity. – 2017-02-17
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    Excellent! Thank you for the explanation. – 2017-02-17

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Your proposed statement is equivalent to saying that for any compact operator $K\in B(L^2(0,1))$, we have $-1\notin\sigma(K)$, which is clearly not true.

To see this, pick an orthonormal basis $\{e_n\}$ for $L^2(0,1)$, let $\{\lambda_n\}$ be a sequence of complex numbers converging to $0$ with $\lambda_1=-1$, and define $K\in B(L^2(0,1))$ by $$K\left(\sum_{n=1}^\infty\alpha_ne_n\right)=\sum_{n=1}^\infty\lambda_n\alpha_ne_n.$$

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    Thanks for the response Aweygan. I see what you're saying, in which case, where exactly does my proof fall apart. In particular, I want to see where my argument is false. – 2017-02-17
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    Additionally, I have edited the question to state that $K$ is a Hilbert-Schmidt integral operator, if that has any bearing. – 2017-02-17
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    @Amit Sorry for not replying sooner. The comments above by D. Thomine address my only concerns with your proof. – 2017-02-17