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I know that a linear operator $T:X \to Y$ (where $X$ and $Y$ are normed vector spaces) is compact if for every sequence $\left(x_{n}\right)\subseteq X$ s.t. $\left\Vert x_{n}\right\Vert \leq C$, the sequence $\left(Tx_{n}\right)\subseteq Y$ has a subsequence which converges in $Y$.

I wish to show that the set of such operators forms a linear subspace of the set of bounded linear operators. In doing so, I have been trying to show that for compact operators $K_1, K_2 : X \to Y$, the operator $K_1 + K_2$ is compact. I think my proof below is on the right track, but I don't think that my forming of the combined subsequence $\left(x_{k, l} \right)$ leads to the desired result (or if it does, it doesn't feel that I've been sufficiently rigorous in the definition of this combined subsequence). I've found a couple of proofs of the statement on the web but they gloss over such details.

Can anyone critique my 'proof' and help me out? Here it is:

Suppose $K_{1},K_{2}:X\to Y$ are compact. Take a sequence $\left(x_{k}\right)$ in $X$ s.t. $\left\Vert x_{k}\right\Vert \leq C$ for each $k\in\mathbb{N}$. Form the sequence $\left(\left(K_{1}+K_{2}\right)x_{n}\right)$. Now there exists subequences $\left(K_{1}x_{n_{k}}\right)$ and $\left(K_{2}x_{n_{l}}\right)$ s.t. $K_{1}x_{n_{k}}\to y_{1}$ and $K_{2}x_{n_{l}}\to y_{2}$ for some $y_{1},y_{2}\in Y$. Hence $$\forall\varepsilon>0:\exists N_{1}:\forall n_{k}\geq N_{1}:\left\Vert K_{1}x_{n_{k}}-y_{1}\right\Vert <\frac{\varepsilon}{2}$$ and $$\forall\varepsilon>0:\exists N_{2}:\forall n_{l}\geq N_{2}:\left\Vert K_{1}x_{n_{l}}-y_{2}\right\Vert <\frac{\varepsilon}{2}.$$ Now form the combined subsequence $\left(x_{k,l}\right)$ by combining the aforementioned subsequences so that their combined indices form a monotonically increasing sequence of integers and pick $N:=\max\left\{ N_{1},N_{2}\right\} .$ Then $\forall\varepsilon>0:\forall n\geq N:$ $$ \left\Vert \left(K_{1}+K_{2}\right)x_{k,l}-\left(y_{1}+y_{2}\right)\right\Vert \leq\left\Vert K_{1}x_{k,l}\right\Vert +\left\Vert K_{2}x_{k,l}\right\Vert <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$

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It might be easier to first take a subsequence for $K_1$, and then take a subsequence of that sequence for $K_2$. – Iian Smythe Apr 29 '13 at 14:30
Yes. Like in the diagonal extraction process, extract from an extraction. Not separately. And when you're done, you should try to show that it is a two-sided ideal, actually, when $X=Y$, in $B(X)$, the bounded linear operaors on $X$. This is an important fact. – 1015 Apr 29 '13 at 14:31
Also, an important fact is that the subspace of compact operators is exactly the (norm) closure of the finite-rank operators. – Iian Smythe Apr 29 '13 at 14:33
@ismythe That's true in Hilbert spaces; it's not true for a general Banach space. – David Mitra Apr 29 '13 at 14:37
@ismythe First counterexample by P. Enflo in 1973: see the approximation property. – 1015 Apr 29 '13 at 14:47
up vote 1 down vote accepted

The subsequences you took must not have any index in commom: For example, let $K_1(x)=K_2(x)=x$ in $\mathbb{C}$ (or $\mathbb{R}$, if you prefer) and take $x_n=(-1)^n$. Then $K_1$ and $K_2$ are compact operators and $\left\{x_n\right\}_n$ is a bounded sequence

If $n_k=2k$ and $n_l=2l+1$, then $\left\{K_1(x_{n_k})\right\}$ and $\left\{K_2(x_{n_l})\right\}$ both converge (to different limits), and there is no subsequence of $\left\{x_n\right\}_n$ that is both a subsequence of $\left\{x_{n_k}\right\}_k$ and $\left\{x_{n_l}\right\}_l$.

Now, let's solve your problem: Let $K_1$, $K_2$ and $\left\{x_n\right\}$ as stated.

First, take a subsequence $\left\{x_{n_k}\right\}_k$ so that $K_1(x_{n_k})$ converges. Then, since $|x_{n_k}|\leq C$, take a subsequence $\left\{x_{n_{k_l}}\right\}_l$ of $\left\{x_{n_k}\right\}_k$ so that $K_2(x_{n_{k_l}})$ converges (you might want to change the notation). Then $\left\{x_{n_{k_l}}\right\}_l$ is a subsequence of the original s.t. $(K_1+K_2)(x_{n_{k_l}})$ converges.

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