bounded operator whose image is contained in the image of a compact operator Let $A,K$ be 2 linear operators from $X$ to $Y$ (Banach spaces). $A$ is bounded and $K$ is compact. If $A(X)\subset K(X)$, is it true that $A$ is alse compact?
I know that if the rank of $K$ is finite, then $A$ must be compact. But I have no idea about the infinite rank case. Any hint would help.
 A: The answer is Yes : the operator $A$ has to be compact. Here is a proof.
For every $n\in\mathbb N$, denote by $M_n$ the closure in $Y$ of the set $K(B(0,n))$, where $B(0,n)$ is the ball with radius $n$ centred at $0$. Then each $M_n$ is compact because $K$ is a compact operator, and obviously $\bigcup_{n\in\mathbb N}M_n$ contains $K(X)$. So we have $A(X)\subseteq \bigcup_{n\in\mathbb N} M_n$.
Now, define $C_n:=\{ x\in X;\; A(x)\in M_n\}$. The $C_n$ are closed subsets of $X$ because the $M_n$ are closed and $A$ is continuous, and $X=\bigcup_{n\in\mathbb N} C_n$. By the Baire category theorem, it follows that one can find $n_0\in\mathbb N$ such that $C_{n_0}$ has nonempty interior. So there is a nontrivial ball $B(x_0,r)$ contained in $C_{n_0}$, i.e. $A(B(x_0,r))\subseteq M_{n_0}$.
From this, one gets by the linearity of $A$ that $A(B(0,r))$ is contained in the translate $M_{n_0}-A(x_0)$, and hence that $A(B(0,1))\subseteq E:=\frac1r\bigl( M_{n_0}-A(x_0)\bigr)$. The set $E$ is compact because $M_{n_0}$ is, and this proves that $A$ is a compact operator.
A: HINT: I would ask if it we're possible for such a compact $K$ to be surjective? Because then you can see immediately it wouldn't be true, however, $K$ cannot be surjective... (why?... hint here) Beyond this, you should note that compact operators may be expressed as operator norm-limits of finite-rank operators, and given that the range of $A$ sits inside the range of $K$, then $A$ should also be a norm-limit of finite rank operators.
