When I read the item of compact operator on Wikipedia, it said that

Let $T_{n}, ~~n\in \mathbb{N}$, be a sequence of compact operators from one Banach space to the other, and suppose that $T_n$ converges to $T$ with respect to the operator norm. Then $T$ is also compact.

Can anyone give me a brief proof of this? Thanks in advance.


A compact operator is one which is the limit (in the norm topology) of finite rank operators. Thus by a diagonal argument, the limit of a norm-convergent sequence of compact operators can be written as the limit of a norm-convergent sequence of finite rank operators, and so is again compact.

Added: As the comment below indicate, this argument is valid in the generality of the question. But perhaps the general approach can be salvaged?

Turning to another characterization of compact operators (valid in full generality), let $T_m \to T$ be our norm convergent sequence of compact operators, and let $x_n$ be a sequence in the unit ball of their domain.

We have to find a Cauchy subsequence of $T(x_n).$ (This will show that $T$ is compact.) We know that any subsequence of $T_m(x_n)$ ($m$ fixed) contains a Cauchy subsequence (since $T_m$ is compact). We know that $T_m(x_n)$ ($n$ fixed) converges to $T(x_n)$, and this convergence is uniform in $n$ (since $|| T(x_n) - T_m(x_n) || \leq || T - T_m|| || x_n|| \leq || T - T_m||,$ which $\to 0$ as $n \to \infty$).

Now a diagonal argument will let us find a Cauchy subsequence of $T(x_n)$.

More precisely, let $x_{n_{1,i}}$ be a subsequence of $x_n$ such that $T_1(x_{n_{1,i}})$ is Cauchy. Pass to subsequences inductively as follows: assuming that we have chosen the subsequence $x_{n_{m,i}}$, take $x_{n_{m+1,i}}$ to be a subsequence of $x_{n_{m,i}}$ such that $T_{m+1}(x_{n_{m+1,i}})$ is Cauchy.

Now define $x_{n_i} = x_{n_{i,i}}$. Then $T_m(x_{n_i})$ is Cauchy for every $m$, and you can deduce from this that $T(x_{n_i})$ is Cauchy.

  • $\begingroup$ got your point!Thanks so much! $\endgroup$ – Craig Jun 10 '13 at 2:05
  • 7
    $\begingroup$ Your first statement is not true for a general Banach space. $\endgroup$ – David Mitra Jun 10 '13 at 2:05
  • 6
    $\begingroup$ The limit of finite rank operators is compact, but not necessarily the other way around (it is true in Hilbert spaces). $\endgroup$ – copper.hat Jun 10 '13 at 2:10
  • $\begingroup$ @DavidMitra: Dear David, Ah, I was a bit worried about this. Thanks for the alert. Regards, $\endgroup$ – Matt E Jun 10 '13 at 2:13
  • $\begingroup$ @copper.hat: Dear Copper, Thanks for pointing this out. Regards, $\endgroup$ – Matt E Jun 10 '13 at 2:14

I think it is simpler to prove this without resorting to sequences.

Key result: In a complete metric space, a set is relatively compact iff it is totally bounded.

Let $B=B(0,1)$ be the open unit ball.

To show that $TB$ is compact, it suffices to show that it is totally bounded.

Choose $\epsilon>0$, and $n$ such that $\|T-T_n\| < \frac{1}{2} \epsilon$. Since $T_n$ is compact, $T_n B$ is totally bounded and hence has a finite $\frac{1}{2}\epsilon$-net $\{t_1,...,t_k \} \subset T_nB$.

I claim that $t_1,...,t_k$ is an $\epsilon$-net for $TB$. Suppose $t \in TB$, then $t=Tx$ for some $x \in B$. Then $\tilde{t} = T_n x$ satisfies $\|t-\tilde{t}\| < \frac{1}{2} \epsilon$. Since $t_1,...,t_k$ is a $\frac{1}{2}\epsilon$-net for $T_nB$, we have $\|\tilde{t}-t_i\| < \frac{1}{2} \epsilon$ for some $i$. Hence $\|t-t_i\| < \epsilon$. Hence $TB$ is totally bounded and so $T$ is compact.


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