How to prove that a bounded linear operator is compact?

Solution 1:

The following proof is adapted from Bruce Barnes, Majorization, range inclusion, and factorization for bounded linear operators. One maybe able to simplify the proof somewhat.

Lemma Let $T,S\in B(X,Y)$ and $R(T)\subseteq R(S)$, then $\exists M > 0$ such that for all $\alpha\in Y^*$, $$ \|T^*\alpha\| \leq M\|S^*\alpha\| \tag{1}$$ where $*$ denotes the adjoint operator.

Proof: Let $U$ be the map from $R(S^*) \to X^*$ given by $U(S^*\alpha) = T^*\alpha$. This map is well-defined since the kernel of $S^*$ is contained in that of $T^*$. It suffices to show that $U$ is a bounded linear operator. Suppose not, then there exists a sequence $\alpha_n$ in $Y^*$ such that $S^*\alpha_n$ has norm 1 and $T^*\alpha_n$ diverges. Now take an arbitrary $x\in X$. By assumption there exists $z\in X$ such that $Sz = Tx$. So $$ T^*\alpha_n(x) = \alpha_n(Tx) = \alpha_n(Sz) = S^*\alpha_n(z) $$ and so $$ |T^*\alpha_n(x)| \leq \|z\| < \infty$$ for each $n$. But by the Uniform Boundedness Principle we have that this implies $$ \sup_n \|T^*\alpha_n\| < \infty $$ and we get a contradiction. q.e.d.

Now recall Schauder's Theorem (Dunford and Schwartz, Chapter VI.5, Theorem 2): An operator is compact if and only if its adjoint is compact.

Corollary If in the previous lemma, $S$ is compact, then so is $T$.

Proof: By Schauder's theorem we have that $S^*$ is compact. Since $T^* = US^*$, and $U$ is a bounded linear operator (with bounded linear extension to $R(S^*)$), we have that $T^*$ is a product of a bounded linear operator with a compact operator, and hence is compact. Appealing to Schauder's theorem again we conclude the proof. q.e.d.

Solution 2:

I found this post when thinking about the same problem. Since people seemed interested for an alternative solution in the end I thought I'd share what I've come up with:

Consider $\overline{K}: X / \text{ker} K \rightarrow \overline{\text{im} K}$. For a bounded sequence $([x_k])_{k \in \mathbb{N}}$ you can find elements $(z_k)_{k \in \mathbb{N}}$ in $\text{ker} K$ such that $$ \Vert x_k - z_k \Vert \leq \sup_{k \in \mathbb{N}} \Vert [x_k] \Vert + 1, $$ so $(x_k - z_k)_{k \in \mathbb{N}}$ is a bounded sequence in $X$. Since $K$ is compact, $(K(x_k - z_k))_k = (\overline{K}([x_k]))_k$ has a convergent subsequence, convergin in $\overline{\text{im} K}$. Thus, $\overline{K}$ is also compact.

Now consider $\tilde{A}: X \rightarrow \text{Im}K$ and $K': X/\text{ker}K \rightarrow F, [x] \mapsto Kx$. We have $A = K' \overline{K}^{-1} \tilde{A}$.

Using the bounded graph theorem, it's easy to show that for Banach spaces $X,Z$, a normed space $Y$, $T: X\rightarrow Y$ linear and bounded and $S: Y \rightarrow Z$ linear and bijective such that $S^{-1}$ is bounded, we have that $ST: X \rightarrow Z$ is bounded.

Applied to our situation, we get that $\overline{K}^{-1} \tilde{A}$ is bounded. Thus, A is compact.