How to prove that a bounded linear operator is compact? 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? 
 A: 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. 
A: 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.
