I'm a little confused about compact operators and whether or not they are invertible. Just hoping someone here can clear up my confusion:

Let $T$ be a compact operator on a Banach space $X$. Since $T$ is compact, we know that $0$ is in the spectrum of $T$. This implies that either $T$ is not invertible, or if it is then the inverse operator isn't bounded. Now, if $T$ did have an inverse $T^{-1}$, then $T$ would be a bijective bounded operator. Hence, by the Banach Open mapping theorem, the map $T^{-1}$ would be bounded. This seems to imply that no compact operator is invertible. But it seems that if $\{x_{n}\}$ is a basis for $X$, then define our operator $T$ so that $Tx_{n}=\lambda_{n}x_{n}$, where, say $\lambda_{n}=2^{-n}$. This operator $T$ is approximated by finite-rank operators, and hence is compact. Also $T^{-1}$ exists, but it obviously isn't bounded.

What is going on here? I feel like I'm going in circles.

  • 2
    $\begingroup$ The operator in your example is not bijective. It is bijective if you consider it from $\mathrm{span}\{x_n\}$ to $\mathrm{span}\{x_n\}$ but not from $\overline{\mathrm{span}} \{x_n\}$ to $\overline{\mathrm{span}} \{x_n\}$. $\endgroup$ – Theo Apr 7 '13 at 21:41

The open mapping theorem is, as you say, about bijective linear operators between Banach spaces. As an operator $T : X \to X$, our $T$ may be injective, but this argument shows it cannot be surjective. If you think of it as an operator $T : X \to \operatorname{ran}(T)$, then it would be bijective, so this argument shows that $\operatorname{ran}(T)$ is not Banach, i.e. not closed in $X$.

When $T$ is injective, $T^{-1}$ is only defined on $\operatorname{ran}(T)$. Part of the confusion may be that when we speak of an "unbounded operator" on $X$, we mean an operator whose domain is a subspace of $X$.


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