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!
[Math] Compact operator whose range is not closed
compact-operatorsfunctional-analysishilbert-spaces
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Best Answer
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.