# Why are these two characterizations of compact operators equivalent?

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 $$T(C_1)$$?

• 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 $y=z$. – 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 $$n_1 such that $$\|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.

• 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. – DanielWainfleet Apr 27 '19 at 2:10
• @DanielWainfleet yes but we only show that the closure of $T(C_1)$ is closed – JustDroppedIn Apr 27 '19 at 8:18