Let $X$ be some Hilbert space. There is the widely known theorem in functional analysis which states that for each closed subspace $H\subset X$ we have $H\bigoplus H^{\perp}=X$.
Now we do not suppose $X$ to be Hilbert space (but it is endowed with a scalar product). Is it true that there always exists such closed subspace $H\subset X$ such that $H\bigoplus H^{\perp}\ne X$?
So, I know that the completness of $X$ is important in the proves given in books, but how to build such closed subspace $H$?
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This question makes no sense because for non Hilbert spaces you do not have definition of $H^\perp\subset X$ – Norbert Oct 17 '13 at 17:55
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@Norbert You only need an inner product (called "scalar product" here, probably because in German it's called Skalarprodukt) to define $H^\perp$. – Daniel Fischer Oct 17 '13 at 17:57
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Yes, I meant $X$ is almost Hilbert, i.e. is not complete (euclidean?) – Oct 17 '13 at 18:02
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Pre-Hilbert is a common name. But note that that does not imply Hausdorffness for all authors. – Daniel Fischer Oct 17 '13 at 18:08
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@DanielFischer I thought OP talked about general Banach spaces – Norbert Oct 17 '13 at 18:28
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@Norbert Is this a duplicate? Surely this question is more general then the one you are linking to? – user1729 Oct 17 '13 at 19:34
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The meaningful question is: Does every closed subspace of a Banach space X have a topologically complementary subspace. The answer to this is: if and only if X is Hilbert.
Edit: I misunderstood the original question. Examples of closed hyperplanes in pre-Hilbert spaces which do not have orthogonal complement are constructed here:
A counterexample to theorem about orthogonal projection. The point is that if a hyperplane $H$ in $X$ has an orthogonal complement $H^\perp=<u>$ ($u$ is the unit vector), then for each vector $v\in X$ one can define the orthogonal projection to $H$ by the usual formula
$$
p(v)= v- (u\cdot v)u.
$$
However, in the linked post one finds examples of closed hyperplanes in pre-Hilbert spaces $X$ so that for some $v\in X$, the orthogonal projection to $H$ does not exist.
Moishe Kohan
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The question, and it is meaningful, is whether in a pre-Hilbert space $X$, we have $X = H \oplus H^\perp$ for all closed subspaces $H \subset X$. – Daniel Fischer Oct 17 '13 at 18:10
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Exactly. I want to understand if it is possible to build such $H$ expicitly (e.g. up to a chose of a Cauchy sequence that has no limit) – Oct 17 '13 at 18:14
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Maybe something like this works?
Starting with $\ell^2(\mathbb{R})$, let $e_i$ be the sequence with a $1$ in the $i$th position, and $0$ elsewhere. Then take $X$ to be the linear span of the $e_i$'s and some sequence $v$, where $v$ has infinitely-many non-zero entries - i.e. the subspace of all (square-summable) sequences with only finitely-many non-zero entries and some sequence like $\left(\dfrac{1}{n}\right)$. Then $X$ is a real inner-product space, but it isn't complete.
Then, if we let $H = \textrm{span}( e_2,\dots,e_n,\dots)$ (i.e. we have left out $e_1$ and $v$), then the orthogonal complement is $\textrm{span}(e_1)$, and $v$ is not in $H$ or $H^\perp$.
It is quite possible I have made a mistake in this answer, but hopefully there is something worthwhile here.
BaronVT
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