## 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\|. 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 $x\in \mathbb{X}$,

$\displaystyle(T^{**}x)(\varphi)=x(T^{*}\varphi)=(T^{*}\varphi)(x)=\varphi(Tx)$

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 $\varphi\in\mathbb{Y}^{*}$,

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

$\Box$

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. $\Box$