For any vector subspace of a Hilbert space the orthogonal complement and the closure of that subspace form a direct sum of the entire space

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Link to it and demote.

• Demote to grade B when a picture of my proof on paper has been included - Alec (talk) 04:07, 8 April 2017 (UTC)

Statement

Let $((X,$$\mathbb{K}$$),\langle\cdot,\cdot\rangle)$ be a Hilbert space and let $L\subseteq X$ be a vector subspace of the vector space $(X,\mathbb{K})$, then:

• $L^\perp$ - the orthogonal complement of $L$ - is a (topologically) closed set of $X$
• $X\eq \overline{L} \oplus L^\perp$ - where $\overline{A}$ denotes the closure of a set $A\in\mathcal{P}(X)$ and $\oplus$ the direct sum of vector spaces
  • This means $\forall x\in X\exists x_1\in\overline{L}\exists x_2\in L^\perp\big[x\eq x_1+x_2\wedge\underbrace{\forall a\in\overline{L}\forall b\in L^\perp[x\eq a+b\implies (a\eq x_1\wedge b\eq x_2)]}_\text{Uniqueness}\ \big]$

Proof

We have some precursors which we use here:

1. The orthogonal complement of a vector subspace is a vector subspace
2. The orthogonal complement of a vector subspace is a topologically closed set
3. The closure of a linear subspace of a normed space is a linear subspace
4. A subspace of a vector space over the real or complex numbers is a convex set and
5. Given a Hilbert space and a non-empty, closed and convex subset then for each point in the space there is a closest point in the subset

Proof:

• Let $x\in X$ be given
  • By given a Hilbert space and a non-empty, closed and convex subset then for each point in the space there is a closest point in the subset we know:

    • $$\exists x_1\in\overline{L}\Big[\Vert x-x_1\Vert\eq\mathop{\text{Inf} }_{a\in \overline{L} }\Big(\Vert x-a\Vert\Big)\Big]$$ and furthermore such an $x_1$ is unique.

  • Define $x_1\in\overline{L}$ to be the unique point such that $\Vert x-x_1\Vert\eq\mathop{\text{Inf} }_{a\in \bar{L} }\big(\Vert x-a\Vert\big)$
    • Now $x_2\eq x-x_1$ is uniquely determined by $x$ and $x_1$ (as a vector space is an (additive) Abelian group)
    • If $x_2\in L^\perp$ we'd be done, we will show this.
      • Suppose $x_2\notin L^\perp$, that means $\exists z\in\overline{L}[\langle x-x_1,z\rangle\neq 0]$
        • Define $z\in\overline{L}$ to be such that $\langle x_2,z\rangle\neq 0$

(Unknown grade)

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See - page 35 (I've annotated it) - I'm not actually sure now if it shows the decomposition is unique though, surely there could be another $x_1'\in\overline{L}$ such that $x_2':\eq x-x_1'\in L^\perp$? I don't think the uniqueness part is shown! Alec (talk) 04:07, 8 April 2017 (UTC)