# Proof of compactness of bounded linear operator

Define $T: l^2 \to l^2$ by $Tx = y =(\eta_j)$, where $x = (\xi_j)$ and $$\eta_j = \sum_{k=1}^{\infty} \alpha_{jk}\xi_k, \quad \quad \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} |\alpha_{jk}|^2 < \infty.$$ Show that $T$ is compact.

Solution:
Let $(T_n)$ be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If $(T_n)$ is uniformly operator convergent, say, $||T_n - T|| \to 0$, then the limit operator $T$is compact. Hence we must show that $||T_n - T|| \to 0$ for $T$ as defined in the problem.

Let $T_n x = (\sum_{k=1}^{n} \alpha_{1k}\xi_k, \sum_{k=1}^{n} \alpha_{2k}\xi_k, \sum_{k=1}^{n} \alpha_{3k}\xi_k, \ldots)$. $T_n$ is linear and bounded. Also, since $\mathrm{dim} \, T_n(X) < \infty$, $T_n$ is compact. Furthermore, $$||(T-T_n)x^2|| = \sum_{j=n+1}^{\infty}\sum_{k=n+1}^{\infty} |\eta_j|^2 = \sum_{j=n+1}^{\infty}\sum_{k=n+1}^{\infty} |\alpha_{jk}|^2 |\xi_k|^2.$$ Now I'm unsure about the last few steps. Hopefully you can help me out. Thanks!

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From your form, You can bound $\|(T - T_n) x\|^2$ by $$\|x\|^2 \sum_{j,k \geq n+1} |\alpha_{jk}|^2$$. Divide by $\|x\|^2$; do you see how this is a bound on $\|T - T_n\|$? Like convergent sums over a single index, convergent sums over double indices vanish as you "exhaust the tail" of the sum, so to speak, so as you take $n \rightarrow \infty$, the sum will go to zero. Now use the closedness property of the space of compact operators. –  A Blumenthal Feb 4 '13 at 4:51
I do! That's brilliant. Since the sum goes to zero as $n$ goes to infinity, we get $||(T-T_n)x|| \to 0$ as $n \to \infty$. Now, since the space of compact operators is closed, we have that the limit of any convergent sequence of compact operators must also be a compact operator, so $T$ is compact. –  Bo Schmidt Feb 5 '13 at 4:09