Possible flaw in "proof" that a sum of two compact operators is compact

Question

If X and Y are Banach spaces, and $A: X \to Y$, $B: X \to Y$ are both compact operators, then $A + B$ is compact.

A + B is compact if and only if for every bounded sequence $\lbrace x_n \rbrace$ in X, the sequence $\lbrace (A + B) x_n \rbrace$ has a convergent subsequence; certainly both $\lbrace A x_n \rbrace$ and $\lbrace B x_n \rbrace$ each have a convergent subsequence, and I've seen a few proofs that seem to think it straightforward that this implies $\lbrace (A + B) x_n \rbrace$ has one. This doesn't seem quite right to me.

The way it appears one should try to make a convergent subsequence is to intersect the indices of the convergent subsequences for A and B individually. So, for example, if $\lbrace A x_{n_j} \rbrace$ and $\lbrace B x_{n_k} \rbrace$, then the idea is to take the set of indices $\lbrace n_m \rbrace := \lbrace n_j \rbrace \cap \lbrace n_k \rbrace$ and say that $\lbrace (A + B) x_{n_m} \rbrace$ converges as a sum of two further subsequences of the convergent subsequences for A and B.

But what if that intersection is empty, for example if $\lbrace n_j \rbrace = 1, 3, 5, ...$ and $\lbrace n_k \rbrace = 2, 4, 6, ...$? Instinctively I would say "if they're disjoint, just pick a different pair", but how can one be sure that the intersection isn't empty for any chosen pair of convergent subsequences?

To be clear, I believe that A + B is compact, it's this form of proof of the fact that seems dubious to me.

Answer 1

If you have two sequences, say $\{a_n\}$ and $\{b_n\}$, and sufficiently neat conditions, then both have a convergent subsequence indexed by the same same set of indices. Formally, I mean that there exists a sequenece of integers $\{n_1<n_2<n_3<\cdots\}$ such that both $\{a_{n_i}\}_{i=1}^\infty$ and $\{b_{n_i}\}_{i=1}^\infty$ both converge.

To guarantee this, you're going to need either

1. each of $\{a_n\}$ and $\{b_n\}$ lie in some compact set (even if they lie if different compact sets, it works); or
2. each of $\{a_n\}$ and $\{b_n\}$ are bounded so you can apply some kind of Bolzano-Weierstrass.

These amount to the same proof though.

In fact, you can get this "common subsequence trick" far more generally: instead of two sequences $\{a_n\}$ and $\{b_n\}$, you can have countably many sequences $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\ldots$. It's the same proof.

For two sequences it's actually very easy to write down. Let $\{a_n\}$ and $\{b_n\}$ be bounded, and say $\{a_n\}$ has convergent subsequence $\{a_{n_i}\}$. But then $\{b_{n_i}\}$ has a convergent subsequence because it is bounded --- say $\{b_{n_{i_k}}\}$ converges. But now $\{a_{n_{i_k}}\}$ is a subsequence of $\{a_{n_i}\}$, hence also convergent. Thus $\{n_{i_1} < n_{i_2} < n_{i_3} < \cdots\}$ is the required set of indices.

For the case with countably many sequences, you can imagine how to proceed. The hardest part is streamlining the notation.

Do you see how to apply this to prove that the sum of two compact operators is compact?

Answer 2

In fact you do it gradually:

Take $x_{n_j}$ such that $Ax_{n_j}$ converges, and then take a sub-subsequence $x_{n_{j_k}}$ such that $Bx_{n_{j_k}}$ converges

Answer 3

Find a convergent subsequence for $A$, then take a sub-subsequence for $B$. You can't necessarily intersect two subsequences, but you can easily take a sub-subsequence.

Answer 4

There are several ways to get rid of this difficulty:

• you can still extract o sequence (for $B$) from an extracted convergent subsequence for $A$.

• or, take an equivalent definition of a compact operator: this is an operator for which the image of a ball is compact.

Answer 5

Consider the following:

We have the sequences $\{Ax_n\},\{B x_n\},$ and $\{(A+B)x_n\}$.

We begin by finding the a convergent subsequence of the first sequence; that is, we find a sequence of indices $\{n_i\}$ such that $\{A_{n_i}\}$ converges. Note that any subsequence of this convergent sequence must also converge.

We now find a convergent subsequence of the sequence $\{B x_{n_i}\}$ over the indices $n_{i_j}$ (which I will simply write as $n_j$, for convenience). Note that both $\{A_{n_j}\}$ and $\{B_{n_j}\}$ converge.

Answer 6

Let $S$ and $T$ be compact operators on a Banach space $X$. Let $(x_n)$ be a bounded sequence in $X$. Then the sequence $(Sx_n)$ has a convergent subsequence call it $(Sx_{n_i})$. Since the sequence $(x_n)$ is bounded it follows that the sequence $(x_{n_i})$ is bounded. Thus $(Tx_{n_i})$ has a convergent subsequence call it $(Tx_{n_{i_j}})$. We also note that since $(Sx_{n_{i_j}})$ is a subsequene of $(Sx_{n_i})$ which is convergent it is also the case that $(Sx_{n_{i_j}})$ is a convergent sequence. It follows from the linearity property for convergent sequences that $([S+T]x_{n_{i_j}})$ is a convergent subsequence of $([S+T]x_n)$ from which we conclude that $S+T$ is compact.