## Unbounded Inverses of Compact Hilbert-Space Operators (Clarification)

I realized this morning that I need to clarify what I mean by “invertible” in yesterday’s post. Suppose $\mathbb{H}$ is a separable Hilbert space and $T: \mathbb{H}\rightarrow\mathbb{H}$ is a compact operator as before. I don’t mean that there exists an operator $S: \mathbb{H}\rightarrow\mathbb{H}$ such that $T\circ S=I, S\circ T=I$. The existence of such an operator is impossible by the Bounded Inverse Theorem. Instead, I mean that $T$ is injective, and therefore $T$ has an inverse latex $S: \mathcal{R}(T)\rightarrow\mathbb{H}$ defined on the range of $T$.

The range of $T$ will be be a proper subspace of $\mathbb{H}$, but it will not be closed unless $T$ is of finite-rank. We know that any finite-dimensional subspace of a Banach space $\mathbb{X}$ is closed. For the reverse direction, we need few lemmas dealing with closed operators. Recall that if $T: \mathcal{D}(T)\rightarrow\mathbb{Y}$ is a linear operator defined in a Banach $\mathbb{X}$ and with range in a Banach space $\mathbb{Y}$, then $T$ is closed if

$(x_{n})_{n=1}^{\infty}\subset\mathcal{D}(T), x_{n}\rightarrow x\in\mathbb{X},Tx_{n}\rightarrow y\in\mathbb{Y}\Longrightarrow Tx_{n}\rightarrow Tx$

Note that the definition of closed operator is different from that of continuous in that we impose the additional hypothesis that the sequence $(Tx_{n})_{n=1}^{\infty}$ converges and don’t a priori require that $x\in\mathcal{D}(T)$.

Lemma 1. Suppose $T: \mathcal{D}(T)\rightarrow\mathbb{Y}$ is a closed linear operator between Banach spaces. If the inverse $T^{-1}$ exists in $\mathbb{Y}$, then $T^{-1}:\mathcal{R}(T)\rightarrow\mathbb{X}$ is closed.

Proof. Suppose $y_{n}=Tx_{n}$ is a sequence in $\mathcal{R}(T)$ such that $y_{n}\rightarrow y\in\mathbb{Y}$ and $x_{n}=T^{-1}y_{n}\rightarrow x\in\mathbb{X}$. Since $T$ is closed, $y=Tx\Longleftrightarrow x=T^{-1}y$. $\Box$

Lemma 2. Suppose $\mathbb{X},\mathbb{Y}$ are Banach spaces and that $T: \mathbb{X}\rightarrow\mathbb{Y}$ is an injective, closed linear operator. Then $T^{-1}: \mathcal{R}(T)\rightarrow\mathbb{Y}$ is bounded if and only if $\mathcal{R}(T)$ is closed.

Proof. First, suppose that $\left\|T^{-1}\right\|<\infty$. Let $y_{n}=Tx_{n}$ be a sequence in $\mathcal{R}(T)$ such that $y_{n}\rightarrow y$. Then $(y_{n})_{n=1}^{\infty}$ is a Cauchy seqence in $\mathbb{Y}$. For $n,m\in\mathbb{Z}^{\geq 1}$,

$\begin{array}{lcl}\displaystyle\left\|x_{n}-x_{m}\right\|=\left\|T^{-1}(Tx_{n})-T^{-1}(Tx_{m})\right\|&=&\displaystyle\left\|T^{-1}\right\|\left\|Tx_{n}-Tx_{m}\right\|\\&=&\displaystyle\left\|T^{-1}\right\|\left\|y_{n}-y_{m}\right\|,\end{array}$

which implies that $(x_{n})_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{X}$. By completeness, $x_{n}\rightarrow x\in\mathbb{X}$. Since $T$ is closed, $y=Tx\in\mathcal{R}(T)$.

Now suppose that $\mathcal{R}(T)$ is closed. Then $\mathcal{R}(T)$ is a Banach space and $T^{-1}: \mathcal{R}(T)\rightarrow\mathbb{X}$ is closed by Lemma 1. We can apply the Closed Graph Theorem, which is equivalent to the Bounded Inverse Theorem, to obtain that $T^{-1}$ is continuous or equivalently, bounded. $\Box$

We have enough tools to show that that $\mathcal{R}(T)$ is necessarily finite-dimensional if $T: \mathbb{X}\rightarrow\mathbb{Y}$ is a compact, linear operator such that $\mathcal{R}(T)$ is a closed subspace of $\mathbb{Y}$. A compact operator is a fortiori closed, and if $\mathcal{R}(T)$ is closed, then $\mathcal{R}(T)$ is a Banach space and $T^{-1}: \mathcal{R}(T)\rightarrow\mathbb{Y}$ is a closed linear operator between Banach spaces by Lemma 1. By Lemma 2, $\left\|T^{-1}\right\|<\infty$, but this contradicts the result of yesterday’s post.