# Compact Hermitian Operator, closed range implies finite rank

Let $K$ be a compact Hermitian operator on a Hilbert space $H$. How to show that if $K$ has infinite rank then the range of $K$ is not closed in $H$?

My thought is that suppose range $K$ is closed in $H$ and then we need to prove that $K$ has finite rank. So far I think that spectral theorem can be applied. I want to find contradiction to compactness, but I still have no idea. Thank you for your help.

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Your idea is the right one, but I would not use the spectral theorem: If the range is closed, the open mapping theorem implies together with compactness that the image of the unit ball is a relatively compact neighborhood of zero, hence the range must be finite-dimensional. You don't need that you deal with a Hermitian operator.

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