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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.

newcworld001
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    Yes. An equivalent formulation of compact operator is that the image under $K$ of closed unit ball in $X$ has compact closure in $Y$. Since the image of the closed unit ball in $X$ under the operator $A$ is a subset of the image of this ball under $K$, we also get the same set inclusion for their closures. Since the closure for $K$ is compact, so too is it such for the closure for $A$. – Sinister Cutlass Dec 08 '15 at 00:56
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    To have $A(B) \subseteq cK(B)$ for the unit ball $B$ and some constant $c>0$ you need an additional argument, e.g. the closed graph theorem applied to $A: X\to K(X)$ where $K(X)$ is endowed with the quotient norm. – Jochen Dec 08 '15 at 08:50
  • @Jochen by the quotient norm you mean the graph norm? i.e, $||x|| + ||Kx||$? – L.F. Cavenaghi May 20 '20 at 02:18
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    @L.F.Cavenaghi No, the quotient norm on the range of an operator $K$ is $|y| =\inf {|x| : K(x)=y}$. – Jochen May 20 '20 at 10:09
  • @Jochen, how do you apply the open mapping theorem if $A$ is not surjective? – L.F. Cavenaghi May 20 '20 at 17:15
  • Where did I say open mapping theorem? You can apply the closed graph theorem. – Jochen May 20 '20 at 18:07
  • Sorry @Jochen, I thought maybe the correct theorem was open mapping theorem, but now I know how to proceed. Clever idea by the way, but not obvious at all. – L.F. Cavenaghi May 20 '20 at 19:29

2 Answers2

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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.

Etienne
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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.

Community
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charlestoncrabb
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    It is not true in general that compact operators are operator norm limits of finite rank operators. It is true, however, in Hilbert spaces. – PhoemueX Dec 08 '15 at 07:37