More on Unbounded Inverses of Compact Operators

In fact, we don’t need to restrict ourselves to injective Hilbert-space operators to show that if T: \mathbb{X}\rightarrow\mathbb{Y} is a compact linear operator between Banach spaces, then \mathcal{R}(T) is closed if and only if \mathcal{R}(T) is finite-dimensional. We know that any finite-dimensional normed space is closed. To prove the reverse direction, we need a few more tools. We begin by proving an elementary yet extremely useful lemma, often called Riesz’s lemma.

Lemma 1. (F. Riesz) Let E be a proper, closed subspace of a Banach space \mathbb{X}. Then for any r\in (0,1), there exists an element y\notin E such that \inf_{x\in E}\left\|y-x\right\|\geq r and \left\|y\right\|=1.

Proof. Let y_{1}\notin E, and set R:=\inf_{x\in E}\left\|y_{1}-x\right\|. By definition of infimum, for any \epsilon>0, there exists x_{1}\in E such that \left\|y_{1}-x_{1}\right\|<R+\epsilon. Set y:=\frac{y_{1}-x_{1}}{\left\|y_{1}-x_{1}\right\|}, so that \left\|y\right\|=1. Then

\begin{array}{lcl}\displaystyle\inf_{x\in E}\left\|y-x\right\|=\inf_{x\in E}\left\|x-\frac{y_{1}-x_{1}}{\left\|y_{1}-x_{1}\right\|}\right\|&=&\displaystyle\dfrac{\inf_{x\in E}\left\|x\left\|y_{1}-x_{1}\right\|-y_{1}+x_{1}\right\|}{\left\|y_{1}-x_{1}\right\|}\\&=&\displaystyle\dfrac{\inf_{x'\in E}\left\|(x'\left\|y_{1}-x_{1}\right\|^{-1}-\left\|y_{1}-x_{1}\right\|^{-1}x_{1})\left\|y_{1}-x_{1}\right\|-y_{1}+x_{1}\right\|}{\left\|y_{1}-x_{1}\right\|}\\&=&\displaystyle\dfrac{\inf_{x'\in E}\left\|x'-y_{1}\right\|}{\left\|y_{1}-x_{1}\right\|}\\&>&\displaystyle\dfrac{R}{R+\epsilon}\end{array}

Varying our choice of \epsilon depending on the given r\in (0,1) completes the proof. \Box

The following lemma is a corollary of Riesz’s lemma which gives a useful criterion for a Banach space to be finite-dimensional.

Lemma 2. Let \mathbb{X} be a Banach space. If every bounded sequence (x_{n})_{n=1}^{\infty} has a convergent subsequence (x_{n_{k}})_{k=1}^{\infty}, then \mathbb{X} is finite-dimensional.

Proof. Suppose \mathbb{X} is infinite-dimensional. By Riesz’s lemma and induction, we can construct a sequence (x_{n})_{n=1}^{\infty} in \mathbb{X} such that \left\|x_{n}\right\|=1 and

\displaystyle\inf_{y \in \text{span}\left\{x_{1},\cdots,x_{n}\right\}}\left\|x_{n+1}-y\right\|\geq\dfrac{1}{2},\indent\forall n\in\mathbb{Z}^{\geq 1}

By hypothesis, (x_{n}) has a convergent subsequence (x_{n_{k}}). But \left\|x_{n_{k+1}}-x_{n_{k}}\right\|\geq \frac{1}{2}. Hence, (x_{n_{k}}) is not Cauchy, which is a contradiction. \Box

For Banach spaces \mathbb{X},\mathbb{Y}, denote their topological duals by \mathbb{X}^{*} and \mathbb{Y}^{*}, respectively.

Theorem 3. A bounded linear operator T:\mathbb{X}\rightarrow\mathbb{Y} is compact if and only if T^{*}:\mathbb{Y}^{*}\rightarrow\mathbb{X}^{*} is compact.

Proof. Suppose that T:\mathbb{X}\rightarrow\mathbb{Y} is compact, and let (\varphi_{k})_{k=1}^{\infty} be a bounded sequence in \mathbb{Y}^{*}. Set M := \sup_{k}\left\|\varphi_{k}\right\|_{\mathbb{Y}^{*}}. Denote the closed unit ball in \mathbb{X} by B. Since T is a compact operator, \overline{T(B)} is a compact subset of \mathbb{Y} and in particular, a Banach space. Define a sequence of functions f_{k}:\overline{T(B)}\rightarrow\mathbb{C} by

\displaystyle f_{k}(y):=\varphi_{k}(y),\indent\forall y\in\overline{T(B)}

I claim that (f_{k})_{k=1}^{\infty} is an equicontinuous, uniformly bounded sequence of functions. Indeed,

\displaystyle\sup_{y\in\overline{T(B)}}\left|f_{k}(y)\right|\leq M\sup_{x\in B}\left\|Tx\right\|=M\left\|T\right\|,\indent\forall k\in\mathbb{Z}^{\geq 1}

and for all y,y' \in\overline{T(B)},

\displaystyle\left|f_{k}(y)-f_{k}(y')\right|=\left|\varphi_{k}(y-y')\right|\leq M\left\|y-y'\right\|_{\mathbb{Y}},\indent\forall k\in\mathbb{Z}^{\geq 1}

By the Arzela-Ascoli theorem, there exists a subsequence (f_{n_{k}})_{k=1}^{\infty} that converges uniformly on \overline{T(B)}. I claim that the sequence (T^{*}\varphi_{n_{k}})_{k=1}^{\infty} is Cauchy in \mathbb{X}^{*}. Indeed,

\begin{array}{lcl}\displaystyle\left\|T^{*}\varphi_{i}-T^{*}\varphi_{j}\right\|_{X^{*}}=\sup_{x\in B}\left|(T^{*}\varphi_{i})(x)-(T^{*}\varphi_{j})(x)\right|&=&\displaystyle\sup_{x\in B}\left|f_{i}(Tx)-f_{j}(Tx)\right|\\&\leq&\displaystyle\sup_{y\in\overline{T(B)}}\left|f_{i}(y)-f_{j}(y)\right|\end{array}

Since \mathbb{X}^{*} is complete, we see that (T^{*}\varphi_{n_{k}})_{k=1}^{\infty} is a convergent subsequence.

Now suppose that T^{*}:\mathbb{Y}^{*}\rightarrow\mathbb{X}^{*} is a compact operator, and let (x_{k})_{k=1}^{\infty} be a bounded sequence in \mathbb{X}. By our first result, T^{**}:\mathbb{X}^{**}\rightarrow\mathbb{Y}^{**} is compct. We know the canonical embeddings of \mathbb{X} into \mathbb{X}^{**} and \mathbb{Y} into \mathbb{Y}^{**} are closed subspaces and therefore Banach spaces. Let (x_{n})_{n=1}^{\infty} be a bounded sequence in \mathbb{X}. Since T^{**} is a compact operator, (T^{**}x_{n})_{n=1}^{\infty} has a convergent subsequence (T^{**}x_{n_{k}})_{k=1}^{\infty}. Then for any continuos functional \varphi\in \mathbb{Y}^{*} and x\in \mathbb{X},


Hence, T^{**}x is the embedding of Tx into \mathbb{Y}^{**}. Thus, the limit of \lim_{k\rightarrow\infty}T^{**}x_{n_{k}}\in\mathbb{Y}\subset\mathbb{Y}^{**}, which implies that \lim_{k\rightarrow\infty}Tx_{n_{k}} exists in \mathbb{Y}. \Box

Lemma 4. Suppose T: \mathbb{X}\rightarrow\mathbb{Y} is a surjective linear map between Banach spaces. Then \mathbb{T}^{*}:\mathbb{Y}^{*}\rightarrow\mathbb{X}^{*} is bounded from below.

Proof. The Open Mapping Theorem for Banach-space operators tells us that there exists c > 0 such that

\displaystyle\left\{y\in\mathbb{Y}:\left\|y\right\|\leq c\right\}\subset \left\{y\in\mathbb{Y}: y= Tx,\left\|x\right\|\leq 1\right\}

By duality, for \varphi\in\mathbb{Y}^{*},

\begin{array}{lcl}\displaystyle\left\|T^{*}\varphi\right\|_{X^{*}}=\sup_{\left\|x\right\|\leq 1}\left|(T^{*}\varphi)x\right|&=&\displaystyle\sup_{\left\|x\right\|\leq1}\left|\varphi(Tx)\right|\\&\geq&\displaystyle\sup_{\left\|y\right\|\leq c}\left|\varphi(y)\right|\\&=&\displaystyle c\sup_{\left\|y\right\|\leq c}\left|\varphi(c^{-1}y)\right|\\&=&\displaystyle c\sup_{\left\|y'\right\|\leq 1}\left|\varphi(y')\right|\\&=&\displaystyle c\left\|\varphi\right\|_{Y^{*}}\end{array}


We now have the preliminary tools and are ready to prove the main result.

If T:\mathbb{X}\rightarrow\mathbb{Y} is a compact operator with closed range \mathcal{R}(T), then \mathcal{R}(T) is finite-dimensional.

Proof. First, suppose that T is bounded from below: there exists a positive constant c such that c\left\|x\right\|\leq\left\|Tx\right\| for all x\in\mathbb{X}. By Lemma 2, it suffices to show that every bounded sequence (Tx_{n})_{n=1}^{\infty} has a convergent subsequence (Tx_{n_{k}})_{k=1}^{\infty}. Since T is bounded from below,

\displaystyle\left\|x_{n}\right\|\leq c^{-1}\left\|Tx_{n}\right\|\leq c^{-1}\sup_{n}\left\|Tx_{n}\right\|,

which implies that (x_{n})_{n=1}^{\infty} is a bounded sequence in \mathbb{X}. Since T is a compact operator, (Tx_{n}) has a convergent subsequence (Tx_{n_{k}}).

Now consder the general case where T is not necessarily bounded from below. Since \mathcal{R}(T) is a closed subspace of \mathbb{Y}, it is a Banach space. Define a compact linear operator S: \mathbb{X}\rightarrow\mathcal{R}(T) by Sx=Tx. By Theorem 1, the dual operator S^{*}:\mathcal{R}(T)^{*}\rightarrow\mathbb{X}^{*} is a compact operator. Since S is surjective, S^{*} is bounded from below by some c>0. Applying our first result shows that S^{*}(\mathcal{R}(T)^{*}) is finite-dimensional. Since S^{*} is an isomorphism (boundedness from below implies injectivity), we see that \mathcal{R}(T)^{*} is finite-dimensional. Since a finite-dimensional space is isomorphic to its dual and \mathcal{R}(T)\hookrightarrow\mathcal{R}(T)^{**}, we conclude that \mathcal{R}(T) is finite-dimensional. \Box

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