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},


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



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.

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One Response to Unbounded Inverses of Compact Hilbert-Space Operators

  1. Pingback: Unbounded Inverses of Compact Hilbert-Space Operators (Clarification) | Math by Matt

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