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Given a general (possibly infinite dimensional) linear vector space $V$ with an inner product, how can you prove that for any subspace $S$, any vector v in $V$ can be uniquely expressed as

$$v = s + t$$

where $s \in S$ and $t$ is orthogonal to $S$?

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    $\begingroup$ You cannot prove that, as this is wrong. Consider $V = \ell^2$ and $S = \{x \in \ell^2 \mid \exists N \;\forall n \ge N: x_n = 0\}$. Then $y \bot S$ implies $y = 0$, hence no $x\in \ell^2 \setminus S$ can be expressed as stated. $\endgroup$
    – martini
    May 21, 2013 at 23:41
  • $\begingroup$ As martini said, your statement isn't quite true. I've given an answer which hopefully is what you were looking for, under a simple additional assumption. $\endgroup$ May 22, 2013 at 9:04

1 Answer 1

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Set $d=\inf_{s\in S} d(v,s)$ to be the distance from $v$ to $S$. We show that finding the nearest point to $v$ lying within $S$ gives the solution.

Lemma 1 There exists $s\in S$ such that $\lVert v-s\rVert = d$ if we additionally assume $S$ complete.

Lemma 2 $t=v-s$ is orthogonal to $S$.

Lemma 3 This decomposition is unique.


Proof of 1 Take $d(s_n,v)\le d^2+\frac 1 n$. We show $s_n$ is Cauchy. Recall the parallelogram law $\lVert v_1+v_2\rVert^2+\lVert v_1-v_2\rVert^2=2(\lVert v_1\rVert^2+\lVert v_2\rVert^2)$. Set $v_1=v-s_n,v_2=v-s_m$. One finds $$\lVert s_m-s_n\rVert^2\le 2/m+2/n$$ and the result follows.

Proof of 2 Let $s'\in S$. $$\lVert v-(s+\lambda s')\rVert^2=\lVert v-s\rVert^2-2 \mathrm{Re}[\left<v-s,s'\right>]\lambda+\lVert s'\rVert^2 \lambda^2$$ but since $s$ minimized the length in question, considering small lambda the linear term must vanish. Hence $v-s$ is orthogonal to $S$.

Proof of 3 $v=s'+t'=s+t$ implies $(s'-s)+(t'-t)=0$. Taking inner products with each term shows both are zero.


As pointed out, for incomplete $S$ this fails. Just choose any situation like that of the finite sequences in $\ell^2$ where the natural 'orthogonal' terms form a series converging to a missing point in $S$.

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