## Unbounded Inverses of Compact Hilbert-Space Operators

I can’t remember anymore, but something compelled me to consult Lokenath’s and Mikusinski’s text Introduction to Hilbert Spaces with Applications and I stumbled upon a cute property of compact operators, which I had forgotten. In case you forget stuff like I do, remember that an operator $T$ on a normed space $\mathbb{X}$ is compact if, for every bounded sequence $(x_{n})_{n=1}^{\infty}$, the sequence $(Tx_{n})_{n=1}^{\infty}$ has a convergent subsequence. The property I had forgotten is that if $\mathbb{H}$ is a separable, infinite-dimensional Hilbert space and $T:\mathbb{H}\rightarrow\mathbb{H}$ is a compact linear operator, then $T^{-1}$ is necessarily unbounded.

When I saw this result stated, I must admit that it took me a while to figure out the proof because I had also forgotten compact operators map weakly convergent sequences to strongly convergent sequences. So, I would like to take this post towards presenting (in an organized fashion), the a proof of the aforementioned property.

We begin by proving the main auxiliary result, which is the weak-to-strong mapping property of compact operators.

Theorem 1. An operator $T$ on a Hilbert space $\mathbb{H}$ is compact if and only if $T$ maps weakly convergent sequences into strongly convergent sequences:

$\displaystyle x_{n}\stackrel{w}\rightarrow x\Longrightarrow Tx_{n}\rightarrow Tx$

Proof. Suppose $T$ is compact and $x_{n}\stackrel{w}\rightarrow x$. If $Tx_{n}\not\rightarrow Tx$, then there exists some $\epsilon > 0$ and a subsequence $(Tx_{n_{k}})_{k=1}^{\infty}$ such that $\left\|Tx_{n_{k}}-Tx\right\|>\epsilon$ for all $k$. Since weakly convergent sequences are bounded by the Banach-Steinhaus theorem, $(Tx_{n_{k}})_{n=1}^{\infty}$ has a strongly convergent subsequence $(Tx_{m_{k}})_{k=1}^{\infty}$. By hypothesis of weak convergence, for every $y \in \mathbb{H}$,

$\displaystyle\langle{Tx_{n},y}\rangle=\langle{x_{n},T^{*}y}\rangle\rightarrow\langle{x,T^{*}y}\rangle=\langle{Tx,y}\rangle,$

which implies that $Tx_{n}\stackrel{w}\rightarrow Tx$ and therefore $Tx_{m_{k}}\stackrel{w}\rightarrow Tx$. Since weak limits are unique and strong convergence implies weak convergence, $Tx_{m_{k}}\rightarrow Tx$. But since $(Tx_{m_{k}})$ is a subsequence of $(Tx_{n_{k}})$, this is a contradiction.

Now suppose that $x_{n}\stackrel{w}\rightarrow x$ implies that $Tx_{n}\rightarrow Tx$. For any bounded sequence $(z_{n})_{n=1}^{\infty}\subset \mathbb{H}$, we need to show that the sequence $(Tz_{n})_{n=1}^{\infty}$ has a convergent subsequence. By hypothesis that $\mathbb{H}$ is separable, it has a complete orthonormal system $(e_{k})_{k=1}^{\infty}$. Let $M>0$ be constant such that $\left\|z_{n}\right\|\leq M$, for all $n$. By Cauchy-Schwarz,

$\displaystyle\left|\langle{z_{n},e_{1}}\rangle\right|\leq M,\indent\forall n$

so there exists a subsequence $(z_{1,n})_{n=1}^{\infty}$ such that the sequence $\langle{z_{1,n},e_{1}}\rangle$ converges. Suppose we have chosen nested subsequences $(z_{1,n})_{n=1}^{\infty}\supset\cdots\supset (z_{k,n})_{n=1}^{\infty}$ such that the sequences $\langle{z_{j,n},e_{j}}\rangle$ converges as $n\rightarrow\infty$. Since $\left|\langle{z_{j,n},e_{j+1}}\rangle\right|\leq M$, there exists a convergent subsequence $(z_{j+1,n})_{n=1}^{\infty}\subset (z_{j,n})_{n=1}^{\infty}$. By induction, we obtained a countable collection of nested subsequences such that the Fourier coefficients $\langle{z_{n,j},e_{j}}\rangle$ converge as $n\rightarrow\infty$.

Define a new sequence $x_{n}:=z_{n,n}$ by taking the diagonal of the subsequences. It is immediate that $\alpha_{k}:=\lim_{n\rightarrow\infty}\langle{x_{n},e_{k}}\rangle$ exists for every $k\in\mathbb{Z}^{\geq 1}$. By Bessel’s inequality, for any $n,l\in\mathbb{Z}^{\geq 1}$,

$\displaystyle\sum_{k=1}^{l}\left|\langle{x_{n},e_{k}}\rangle\right|^{2}\leq\sum_{k=1}^{\infty}\left|\langle{x_{n},e_{k}}\rangle\right|^{2}=\left\|x_{n}\right\|^{2}\leq M^{2}$

Letting $n\rightarrow\infty$, we obtain

$\displaystyle\sum_{k=1}^{l}\left|\alpha_{k}\right|^{2}\leq M^{2}\Longrightarrow\sum_{k=1}^{\infty}\left|\alpha_{k}\right|^{2}\leq M^{2}$

An application of Parseval’s indetity yields that the series $\sum_{k=1}^{\infty}\alpha_{k}e_{k}$ is convergent in $\mathbb{H}$, and we denote its limit by $z$. I claim that $\langle{x_{n},e_{m}}\rangle\rightarrow\langle{z,e_{m}}\rangle$, for every $m\in\mathbb{Z}^{\geq 1}$, as $n\rightarrow\infty$. Indeeed, this follows immediately from observing that

$\begin{array}{lcl}\displaystyle\langle{x_{n},e_{m}}\rangle-\langle{z,e_{m}}\rangle&=&\displaystyle\langle{x_{n},e_{m}}\rangle-\left\langle{\sum_{k=1}^{\infty}\alpha_{k}e_{k},e_{m}}\right\rangle\\&=&\displaystyle\langle{x_{n},e_{m}}\rangle-\alpha_{m}\rightarrow 0\end{array}$

as $n\rightarrow\infty$. Since ${\text{span}\left\{e_{k}\right\}_{k=1}^{\infty}}$ is dense in $\mathbb{H}$, it follows that $x_{n}\stackrel{w}\rightarrow z$. Hence, $Tx_{n}\rightarrow Tz$. Recalling that $(Tx_{n})_{n=1}^{\infty}$ is a subsequence of $(Tz_{n})_{n=1}^{\infty}$ completes the proof. $\Box$

Cor 2. Let $\mathbb{H}$ be as above. If $T:\mathbb{H}\rightarrow\mathbb{H}$ is compact any $(e_{n})_{n=1}^{\infty}$ is an orthonormal sequence, then $Te_{n} \rightarrow 0$.

Proof. We only need to show that an orthonormal sequence $e_{n}\stackrel{w}\rightarrow 0$. By Bessel’s inequality, $\left\|x\right\|^{2}\geq\sum_{k=1}^{\infty}\left|\langle{x,e_{k}}\rangle\right|^{2}$, which implies that

$\displaystyle\lim_{k\rightarrow\infty}\left|\langle{x,e_{k}}\rangle\right|=0$

$\Box$

An operator $A$ defined on a Hilbert space $\mathbb{H}$ is said to be unbounded, if the range $\mathcal{R}(A) \subset\mathbb{H}$ is not a bounded subset. Since $A$ is bounded if and only if $A$ is continuous at the origin $0$, an operator $A$ is unbounded if and only if there exists a sequence $x_{n}\rightarrow 0$ but $Ax_{n}\not\rightarrow 0$, as $n \rightarrow\infty$.

The differential operator on $L^{2}(\mathbb{R})$ is a standard example of an unbounded operator; we will prove this result in a later post. More generally, any compact, invertible operator $A$ on an infinite-dimensional (separable) Hilbert space $\mathbb{H}$ has an unbounded inverse $A^{-1}$. Indeed, $\mathbb{H}$ necessarily has an orthonormal sequence $(e_{n})_{n=1}^{\infty}$. For each $n$, there exists a unique $v_{n}\in\mathbb{H}$ such that $Ae_{n}=v_{n}\Longleftrightarrow e_{n}=A^{-1}v_{n}$. Since $e_{n}\stackrel{w}\rightarrow 0$ and $A$ is compact, $v_{n}\rightarrow 0$. But $A^{-1}v_{n}\not\rightarrow 0$, since $\left\|A^{-1}v_{n}\right\|=1$ for all $n$.