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},


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.

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