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I encountered a homework problem that says:

If $A$ is a bounded linear operator from $X$ to $Y$. And $K$ is a compact operator from $X$ to $Y$, where $X$ and $Y$ are both Banach spaces, and Ran$(A)\subset$ Ran$(K)$. Then $A$ is also a compact operator.

I tried to use the definition of a compact operator to solve this one. (Indeed, the professor only covered the definition of compact operator in class and said that it would be enough for the homework problems.) Here's what I did. I started by choosing a bounded sequence $x_n$ in X and since A is bounded, $A(x_n)$ is also bounded. And from the assumption that $R(A)\subset R(K)$, I conclude that there exist $y_n\in X$, s.t. $K(y_n)=A(x_n)$. Now if I can somehow prove that $y_n$ is bounded in X, I can easily prove the problem by using the compactness of K. But this is exactly the place where I am stuck. Please help me out. Am I going along the right path?

Also, I had another problem saying that: If X is infinitely dimensional and K is an compact operator and is one to one, then I-K must not be compact.

I proved this one, but didn't really use the assumption that K is one to one. I looked over and over again but couldn't nd where I made the mistake.

Here's what I did: Choose any sequence in X that is of norm 1. Then suppose I-K is compact. It follows there must exists a subsequence $x_{n_k}$ that $(I-K)(x_{n_k})$ converges. And since K is compact, there exists a sub-subsequence $x_{n_{k_j}}$ that $K(x_{n_{k_j}})$. Now I claim that in fact $x_{n_{k_j}}$ converges in X. Indeed, $x_{n_{k_j}}=(I-K)(x_{n_{k_j}})+K(x_{n_{k_j}})$. Hence, for any sequence on the unit sphere, I've found a subsequence that converges. This means the unit sphere is compact, which contradicts with X being infinitely dimensional.

Did I do something wrong?

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There's nothing wrong with your solution to the second problem and the one-to-one condition is unnecessary. – minimalrho Mar 13 '12 at 22:24
@minimalrho, thank you so much. Do you have any clues for the first one? – henryforever14 Mar 13 '12 at 22:47
The first problem seems tricky. Note, though, that trying to prove $y_n$ is bounded will not help, because $y_n$ need not be bounded. (For example, if $K$ is not one-to-one, $y_n$ could have an arbitrarily large component in the kernel of $K$.) – Nate Eldredge Mar 13 '12 at 23:25
@NateEldredge That is true. But somehow, I need only $d(y_n,ker(K))$ to be bounded. And that is enough to prove this problem. – henryforever14 Mar 13 '12 at 23:31
Hmm: you need $X$ to be Banach; here's why. Let $X=c_{00}$ and $Y=c_0$, both with the supremum norm, and let $A$ be the inclusion map, which is not compact. Let $K(e_n)=e_n/n$ where $e_n$ are the usual basis vectors. Then $K$ extends continuously to $c_0$ (the completion of $X$) and $K$ is compact. However, clearly the ranges of $A$ and $K$ are equal (both just $c_{00}\subseteq c_0$). So "metamathematically" you are going to have to use one of the tools from Banach Space theory, as the result fails for normed spaces. – Matthew Daws Mar 14 '12 at 16:56
up vote 4 down vote accepted

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.

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I think their professor gone crazy if he wanted to get this solution using only definition given on classes. – Norbert Mar 14 '12 at 17:30
@Willie Wong how is the U constructed? – henryforever14 Mar 14 '12 at 20:19
@Henry Since the kernel of $S^*$ is contained in that of $T^*$ we just define using the expression $US^* = T^*$. – Willie Wong Mar 15 '12 at 9:20
@Norbert: yes. I think so too. It is possible that this proof can be simplified (in particular the use of Schauder's Theorem may be a bit gratuitous). The key is to construct something like $U$ and show its boundedness using some version of the uniform boundedness principle, and I think that part probably is necessary. – Willie Wong Mar 15 '12 at 9:22
@Norbert: the natural thing to try, I guess, after seeing the above, is whether we can start by constructing $V$ such that $SV = T$, where $V$ is defined from $X/\ker{T}\to X$, and showing directly that $V$ is bounded. But I don't know whether it will be much simpler than the above, if it works. – Willie Wong Mar 15 '12 at 9:26

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 $\overline{K}: X / \text{ker} K \rightarrow \text{im} K$. 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. Thus, $\overline{K}$ is also compact.

Now consider $\tilde{A}: X \rightarrow \text{Im}K$. We have $\tilde{A} = \overline{K} \overline{K}^{-1} \tilde{A}$.

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, $\tilde{A}$ and also A (by using the sequence criterion again) is compact.

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