If $T$ is a bounded linear operator on a Hilbert space $H$, then I have heard that the following two things are true:

  1. $T$ is compact if and only if $T(C_1)$ is compact where $C_1$ is the closed unit ball.
  2. $T$ 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 $T(C_1)$?

  • $\begingroup$ You can use the following: A bounded sequence is a Hilbert space has a weakly convergent subsequence. Bounded operators are weak-weak continuous. If $(y_n) $ converges weakly to $y$ and in norm to $z$, then $y=z$. $\endgroup$ – David Mitra Apr 26 '19 at 4:56

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.

  • $\begingroup$ If $T(C_1)$ is compact then $T(C_1)$ is closed because we are in a Hausdorff space, where compact sets are necessarily closed. $\endgroup$ – DanielWainfleet Apr 27 '19 at 2:10
  • $\begingroup$ @DanielWainfleet yes but we only show that the closure of $T(C_1)$ is closed $\endgroup$ – JustDroppedIn Apr 27 '19 at 8:18

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