Compact operator whose range is not closed

Question

I am asked to find a compact operator (on a Hilbert space) whose range is not closed, but I am having trouble coming up with one. My guess is that you need to have some sequence in the range that converges to something outside the range, but this feels like it contradicts the definition of compactness. [Our definition of an operator being compact is that the closure of the image of the closed unit ball is sequentially compact.] Any pointers/suggestions would be appreciated. Thanks in advance!

Answer 1

In fact, the range of a compact operator is never closed unless the operator has finite rank.

The example $T:(x_n)\mapsto (x_n/n)$ is standard. Observe that the range contains all finite sequences, and therefore is dense. If it was also closed, it would be everything. But the element $(1/n)$ is not in the range.

Just as a remark: for $T$ as above, the image of the closed unit ball is in fact closed (and compact). But the image of the whole space is not. Taking the countable union of images of balls of various radii, we lose closedness.

Answer 2

As another example:

If we let $L(D)$ denote the space of holomorphic functions on a disk $D$, and if $D_1 \subset D_2$ is an inclusion of the disk of radius $r_1$ (around $0$, say) in the disk of radius $r_2$, with $r_1 < r_2$, then restriction $L(D_1) \to L(D_2)$ is compact with dense image.

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Indeed, suppose that $B_1 \to B_2$ is a compact operator between Banach spaces with closed image. Replacing $B_2$ by the image, we may assume it is surjective. Then, by the open mapping theorem, it is open. But then $B_2$ has a compact neighbourhood of its origin, and so is finite dimensional. (As was noted in the other answer.)

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In practice, then, compact operators often have dense image: like restricting from a bigger disk to a smaller one, or mapping one Sobolev space into another.