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I'm working through some notes for my signal processing class and they introduce the whole notion of pre-Hilbert spaces (inner product spaces) essentially only in order to be able to project elements onto closed subspaces.

My question is: suppose $V$ is a general normed space (not necessarily derived from an inner product) and $U \subset V$ is a complete subset, can we not use the same proof to show that $\forall v \in V, \exists u \in U$ such that $||u-v||$ is minimal? In that case, the only thing the inner product does is to make the projection orthogonal, right?

Othman El Hammouchi
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  • It’s much more complicated if you are not in a Hilbert space. You do not have an algorithm to minimise that norm. You should compute its derivative, provided it exists, set it equal to zero, etc, etc... Plus, the projection might not be unique. I am not even sure that it always exists. There is a famous big problem in functional analysis called the “complemented subspace problem”, which is concerned exactly with this question, if I recall correctly. That’s a massive can of worms. – Giuseppe Negro Mar 09 '21 at 11:06
  • I might be massively mistaken here, but can't you just modify the standard proof of this for inner product spaces? You find a sequence of vectors in the subspace s.t. their distance to converges to the infimum, prove that this is Cauchy and hence derive the existence of a limit within the subspace. – Othman El Hammouchi Mar 09 '21 at 11:13
  • It is a well known fact that every closed convex subset of a normed space X is proximinal if and only if X is reflexive. – Evangelopoulos Foivos Mar 09 '21 at 12:31
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    Let us know when you've done your "can't you just do this" stuff – David C. Ullrich Mar 09 '21 at 15:14

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You can always project onto any subspace at all. Namely, given any $W\leq V$, you can choose an algebraic basis $B_1$ of $W$ and extend it to a basis $B=B_1\sqcup B_2$ of $V$. Then for any $v\in V, v=\sum_{b_1\in B_1} a_{b_1}b_1+\sum_{b_2\in B_2}a_{b_2}b_2$, you can put $F(v)=\sum_{b_1\in B_1} a_{b_1}b_1$, which gives you a projection $V\to W$, and any projection is of this form.

If $V$ is a topological vector space, e.g. a normed space, then such a projection will typically not be continuous, however, and $F(v)$ may not minimize the distance from $v$ to $W$ (even if $F$ happens to be continuous with respect to some given metric). In general, there may be no continuous projection at all, not even if $W$ is a closed (or complete, when it makes sense) subspace of $V$.

If $V$ is an arbitrary normed space, and you try to mimic the proof you mention, it does does not work: given any $v\in V$, even if you choose a sequence $w_n\in W$ such that $\lvert v-w_n\rvert<1/n+\inf_{w\in W}\lvert v-w\rvert$, the sequence $w_n$ need not be Cauchy. For example, if $V$ is the plane with the $L^\infty$ norm and $W$ is spanned by $(0,1)$, then for $v=(1,0)$, you can choose $w_n=(0,(-1)^n)$ and this is clearly not Cauchy, even though $\lvert v-w_n\rvert=1=\inf_{w\in W}\lvert v-w\rvert$. In this case, a distance-minimizing projection does exist, but it is not unique.

tomasz
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