If $T$ is a bounded linear operator on a Hilbert space $H$, then I have heard that the following two things are true:
- $T$ is compact if and only if $T(C_1)$ is compact where $C_1$ is the closed unit ball.
- $T$ is compact if and only if for every bounded sequence $(x_n)$, $T(x_n)$ has a convergent subsequence.
My question is, why are these two characterizations of compact operators on Hilbert spaces equivalent? Specifically, why does the the second clause of statement 2 imply the second clause of statement 1?
If for every bounded sequence $(x_n)$, $T(x_n)$ has a convergent subsequence, why does that imply $T(C_1)$ is compact? After all, couldn't it be there be a sequence $(x_n)$ in $C_1$ such that $(T(x_n))$ has subsequences which converge to points in $H$, but none of those subseqeunces converge to a point in $T(C_1)$?
Best Answer
In general, if $X,Y$ are Banach spaces and $T$ is a linear operator from $X$ to $Y$, then the following are equivalent:
(i) For every bounded sequence $(x_n)\subset X$ there exists a convergent subsequence of $(Tx_n)\subset Y$
(ii) For every bounded set $E\subset X$ the set $T(E)$ has compact closure.
The implication $(ii)\implies (i)$ is obvious (consider $E=\{x_n:n\in\mathbb{N}\}$ for a bounded sequence $(x_n)\subset X$), so we will show the implication $(i)\implies (ii)$:
Let $E\subset X$ be a bounded set and let $(y_n)\subset \overline{T(E)}$. Now for each $n\in\mathbb{N}$ there exists $(x_m^{(n)})_{m=1}^\infty\subset E$ such that $Tx_m^{(n)}\to y_n$, as $m\to\infty$. So, for $n=1$ we can find $n_1$ such that $\|Tx_{n_1}^{(1)}-y_1\|<1$. For $n=2$, we can find $n_2>n_1$ such that $\|Tx_{n_2}^{(2)}-y_2\|<1/2$. So in general we find indexes $n_1<n_2<\dots$ such that $\|Tx_{n_k}^{(k)}-y_k\|<1/k$ for all $k\in\mathbb{N}$. We consider the sequence $(x_{n_k}^{(k)})_{k=1}^\infty\subset E$ which is bounded (since $E$ is bounded). By assumption $(i)$, there exists a subsequence of indexes $(k_l)$ such that $Tx_{n_{k_l}}^{(k_l)}\to y\in\overline{T(E)}$ for some $y\in\overline{T(E)}$. Now we will show that $y_{k_l}\to y$: it is $$\|y_{k_l}-y\|\leq\|y_{k_l}-Tx_{n_{k_l}}^{(k_l)}\|+\|Tx_{n_{k_l}}^{(k_l)}-y\|\leq1/k_l+\|Tx_{n_{k_l}}^{(k_l)}-y\|\to0$$ and we are done.
To fully answer your question, it still remains to show that $T(C_1)$ is closed. I suppose that the Hilbert structure comes in here.