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I am having a chicken egg problem with projections in Hilbert spaces. I was trying to show that if a Hilbert space can be written as $H = U \oplus U^\bot$ where $U$ is any subspace then $U$ is closed.

In an answer it was pointed out to me that I can argue that $U$ is the range of a projection. But my understanding is that if $U$ is closed then $H=U \oplus U^\bot$ and the map $u + u^\bot \mapsto u$ is a continuous linear operator and we call it the ortogonal projection onto $U$.

But in the answer the suggestion goes the other way around: we don't have that $U$ is closed.

(1) If $U$ is not closed is it still true that every $h \in H$ can be written as $u + u^\bot$ and we still get an orthogonal projection?

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(2) It's not clear whether I should projection onto $U$ or onto $U^\bot$. This seems to suggest that both the range and the kernel of a projection are always closed. Is this really the case?

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1 Answers1

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I think the easiest way to approach this question as follows: suppose $u\in\overline{U}$. Then $u\in H$, so $u=v+w$ for some uniquely defined $v\in U$, $w\in U^\perp$. Then $u-v=w\in\overline{U}\cap U^\perp$, so since $\overline{U}=U^{\perp\perp}$, we have $u-v=w=0$. In particular, $u=v\in U$. Hence $U=\overline{U}$ is closed.

In answer to question (1), no, you cannot write any $h\in H$ as $H=u+u^\perp$ for some $u\in U,u^\perp\in U^\perp$ if $U$ is not closed. For any subset $U$ we have $U\cap U^\perp\subseteq0$, so if we had $H=U+U^\perp$ then the sum would be direct. As we have just seen, this would imply $U$ is closed.

Jason
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  • It's not clear to me whether this is an answer to question (1) or question (2). –  Feb 09 '15 at 00:45
  • It's an answer to the original question which avoids any notion of projection, which seemed to confuse you. So I guess my answer to the second question is "neither". I will add in an answer to question 1. – Jason Feb 09 '15 at 00:47