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Let $A \in L(X,Y)$ be a linear operator between Hilbert spaces and the operator $$\hat{A}: \ker(A)^{\perp} \rightarrow \operatorname{ran}(A)$$ is a restriction of $A$ which is bijective. Now $\ker(A)^{\perp}$ is certainly a closed space and since this operator is bijective and continuous, I would also suspect that this means that $\operatorname{ran}(A)$ is closed, because $\hat{A}$ is an isomorphism. But of course not every range of an operator is closed, so there has to be a mistake in this argument.

Edit: Ah got it. The inverse operator does not need to be bounded, as the open mapping theorem only works if $\operatorname{ran}(A)$ is closed, too.

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  • $\begingroup$ You can post an answer to your own question, for the benefit of everyone :) $\endgroup$ – Zev Chonoles Jun 26 '15 at 0:37
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The inverse operator does not need to be continuous, so we cannot conclude that the range is closed.

In fact, if the range is closed, this immediately implies that the inverse operator is continuous by the open mapping theorem.

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