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?
Best Answer
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.