orthogonal decomposition theorem
Theorem  Let $X$ be an Hilbert space^{} and $A\subseteq X$ a closed subspace. Then the orthogonal complement^{} (http://planetmath.org/Complimentary) of $A$, denoted ${A}^{\u27c2}$, is a topological complement of $A$. That means ${A}^{\u27c2}$ is closed and
$$X=A\oplus {A}^{\u27c2}.$$ 
Proof :

•
${A}^{\u27c2}$ is closed :
This follows easily from the continuity of the inner product^{}. If a sequence^{} $({x}_{n})$ of elements in ${A}^{\u27c2}$ converges to an element ${x}_{0}\in X$, then
$$\u27e8{x}_{0},a\u27e9=\u27e8\underset{n\to \mathrm{\infty}}{lim}{x}_{n},a\u27e9=\underset{n\to \mathrm{\infty}}{lim}\u27e8{x}_{n},a\u27e9=0\text{for every}a\in A$$ which implies that ${x}_{0}\in {A}^{\u27c2}$.

•
$X=A\oplus {A}^{\u27c2}$ :
Since $X$ is complete^{} (http://planetmath.org/Complete) and $A$ is closed, $A$ is a subspace^{} of $X$. Therefore, for every $x\in X$, there exists a best approximation of $x$ in $A$, which we denote by ${a}_{0}\in A$, that satisfies $x{a}_{0}\in {A}^{\u27c2}$ (see this entry (http://planetmath.org/BestApproximationInInnerProductSpaces)).
This allows one to write $x$ as a sum of elements in $A$ and ${A}^{\u27c2}$
$$x={a}_{0}+(x{a}_{0})$$ which proves that
$$X=A+{A}^{\u27c2}.$$ Moreover, it is easy to see that
$$A\cap {A}^{\u27c2}=\{0\}$$ since if $y\in A\cap {A}^{\u27c2}$ then $\u27e8y,y\u27e9=0$, which means $y=0$.
We conclude that $X=A\oplus {A}^{\u27c2}$. $\mathrm{\square}$
Title  orthogonal decomposition theorem 

Canonical name  OrthogonalDecompositionTheorem 
Date of creation  20130322 17:32:34 
Last modified on  20130322 17:32:34 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  4 
Author  asteroid (17536) 
Entry type  Theorem 
Classification  msc 46A99 
Synonym  closed subspaces of Hilbert spaces are complemented 