# Compact Operator with Infinite rank Doesn't have a Closed Image

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The trick is to show the contrapositive. If an operator $T$ has infinite-dimensional and closed image, then it is not compact. Indeed, by restriction we get a bijective bounded linear operator $T:(\ker T)^\perp\to\text{Im}\,T$. By the open mapping theorem, $T$ maps open sets to open sets. So the image of the unit ball is open, and this is not a compact subset of $\text{Im}\,T$ because $\text{Im}\,T$ is an infinite-dimensional Hilbert space.

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### PPR

Updated on August 01, 2022

• PPR 3 months

Let $\mathcal{H}$ be a separable infinite-dimensional Hilbert space.

Claim: A compact operator $T$ which has infinite-rank has an image that isn't closed.

I'm trying to prove this claim but I'm faced with some difficulties.

Here is what I attempted:

Proof:

Without loss of generality, because $T$ is compact it is a norm limit of finite-rank operators, so that we may write $T$ as the following norm-limit:

$$T = \lim_{N\to\infty}\sum_{n=1}^{N}\alpha(n) u(n) \left<v(n),\,\cdot\right>$$

where $u:\mathbb{N}\to\mathcal{H}$ and $v:\mathbb{N}\to\mathcal{H}$ are two orthonormal bases of $\mathcal{H}$, and $\alpha:\mathbb{N}\to\mathbb{C}$ is some bounded sequence of complex numbers. Without loss of generality, $\alpha$ is never zero. Indeed, by the fact that $T$ has infinite rank, we know that infinitely many points of $\mathbb{N}$ map to non-zero values under $\alpha$. Then we could just as well exclude all those (finite) points of $\mathbb{N}$ for which $\alpha$ is zero and obtain a closed subspace of $\mathbb{H}$ to work with.

Next, $\alpha$ has to be bounded because $$||T|| = \sup\left(\{\left|\alpha(n)\right|\,|\,n\in\mathbb{N}\}\right)$$Indeed, assume $\alpha$ is not bounded. Then $||T||$ is not finite, because for any given number $M>0$, we could find some $n_M\in\mathbb{N}$ such that $\left|\alpha(n_M)\right|>M$. Then \begin{align} ||T||& \equiv \sup\left(\{||T\psi||\,|\,\psi\in\mathcal{H}\,:\,||\psi||=1\}\right)&\\ &\geq ||Tv(n_M)|| &(||v(n_M)||=1)\\&=|\alpha(n_M)|&\\&>M \end{align}

We conclude that $\sup\left(\{\left|\alpha(n)\right|\,|\,n\in\mathbb{N}\}\right)$ must be finite to be compatible with $||T||$ finite.

Given that $\sup\left(\{\left|\alpha(n)\right|\,|\,n\in\mathbb{N}\}\right)$ is finite, we have just seen that $$||T||\geq\sup\left(\{\left|\alpha(n)\right|\,|\,n\in\mathbb{N}\}\right)$$ and in fact due to \begin{align} ||T\psi||&=||\sum_{n=1}^{\infty}\alpha(n)u(n)\left<v(n),\,\psi\right>||\\&\leq \sum_{n=1}^{\infty}|\alpha(n)||\left<v(n),\,\psi\right>|\underbrace{||u(n)||}_{=1}\\&\leq\sup\left(\{\left|\alpha(n)\right|\,|\,n\in\mathbb{N}\}\right)\underbrace{\sum_{n=1}^{\infty}|\left<v(n),\,\psi\right>|}_{=||\psi||}\end{align} so that really $$||T||=\sup\left(\{\left|\alpha(n)\right|\,|\,n\in\mathbb{N}\}\right)$$

Next, we derive a useful condition for a vector to lie in $im(T)$:

\begin{align}\psi\in im(T)&\Leftrightarrow\exists\varphi\in\mathcal{H}:T\varphi=\psi\\ &\Leftrightarrow\exists\varphi\in\mathcal{H}:\alpha(n)\left<v(n),\,\varphi\right>=\left<u(n),\,\psi\right>\forall n\in\mathbb{N}\\ &\Leftrightarrow\sum_{n=1}^{\infty}\left|\frac{\left<u(n),\,\psi\right>}{\alpha(n)}\right|^2<\infty\end{align}

So the strategy now would be to define a vector outside of $im(T)$: $$\psi := \sum_{n=1}^{\infty}\alpha(n) \beta(n) u(n)$$ for some sequence $\beta:\mathbb{N}\to\mathbb{C}$ such that $\alpha\cdot\beta$ is in $l^2(\mathbb{N})$ yet $\beta$ is NOT in $l^2(\mathbb{N})$, and further, define a sequence $\varphi:\mathbb{N}\to im(T)$ by $$\varphi(m) := \sum_{n=1}^{\infty}\alpha(n) \beta(n) \gamma(m,n) u(n) \,\forall m\in\mathbb{N}$$ for some sequence $\gamma:\mathbb{N}^2\to\mathbb{C}$ such that $$\lim_{m\to\infty}\gamma(m,n)=1$$ for all $n$ and such that $\alpha\cdot \beta\cdot \gamma(m,\cdot)$ would lie in $l^2(\mathbb{N})$ AND $\beta\cdot \gamma(m,\cdot)$ also would lie in $l^2(\mathbb{N})$ for all $m\in\mathbb{N}$.

Then we would have $\lim_{m\to\infty}\varphi(m) = \psi$ which proves $$im(T)\notin Closed\left(\mathcal{H}\right)$$

Question: Given such sequence $\alpha$, how to prove that the sequences $\beta$ and $\gamma$ with the required properties exist? I know how to do it for $\alpha(n) = \frac{1}{n}$, but I'm looking for a general formulation. Alternatively, is there a better way to show the claim?

• reuns over 6 years
(say all the $\alpha(n)$ are $\ne 0$) $|\alpha(n)| \to 0$ when $n \to \infty$, otherwise $T$ is not compact. hence, $\frac{1}{|\alpha(n)|} \to \infty$ and there are vectors $\psi = \sum_n c(n) u(n)$, i.e. $\sum_n |c(n)|^2$ converges, such that $\sum_n \left|\frac{c(n)}{\alpha(n)}\right|^2$ diverges
• PPR over 6 years
@user1952009, how do you show that $T$ compact implies $|\alpha(n)|\to0$? I only concluded that $\alpha$ is bounded.
• reuns over 6 years
simply that $T v(n)$ doesn't have a convergent subsequence
• PPR over 6 years
Thanks for your answer. I feel like a bit more elaboration is necessary (at least for me). Denoting by $S:(\ker(T))^\perp\to im(T)$ the restriction, we know that $T$ cpt leads to $S$ cpt. Next, $S$ is an open mapping, so if $U$ is the open unit ball in $(\ker(T))^\perp$, $S(U)\in Open(im(T))$. Now as $S$ is compact, we should have $\bar{S(U)}$ cpt in $im(T)$, which it cannot be because $im(T)$ is infinite dimensional (elaborate how to show this using Riesz' lemma..?)
• Martin Argerami over 6 years
Yes, that's it. I'm taking it as a basic fact that in an infinite-dimensional normed vector space balls are not compact. In a Banach space you would need Riesz' Lemma. But in a Hilbert space, if you have a ball around the origin you can put an infinite orthonormal set inside; the tips of these vectors give you a sequence (in the separable case) that does not admit a convergent subsequence.