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orthogonal decomposition theorem
Theorem  Let $X$ be an Hilbert space and $A\subseteq X$ a closed subspace. Then the orthogonal complement of $A$, denoted $A^{{\perp}}$, is a topological complement of $A$. That means $A^{{\perp}}$ is closed and
$X=A\oplus A^{{\perp}}\;.$ 
Proof :

$A^{{\perp}}$ is closed :
This follows easily from the continuity of the inner product. If a sequence $(x_{n})$ of elements in $A^{{\perp}}$ converges to an element $x_{0}\in X$, then
$\langle x_{0},a\rangle=\langle\lim_{{n\rightarrow\infty}}x_{n},a\rangle=\lim_{% {n\rightarrow\infty}}\langle x_{n},a\rangle=0\;\;\;\text{for every}a\in A$ which implies that $x_{0}\in A^{{\perp}}$.

$X=A\oplus A^{{\perp}}$ :
Since $X$ is complete and $A$ is closed, $A$ is a complete 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 $xa_{0}\in A^{{\perp}}$ (see this entry).
This allows one to write $x$ as a sum of elements in $A$ and $A^{{\perp}}$
$x=a_{0}+(xa_{0})$ which proves that
$X=A+A^{{\perp}}\;.$ Moreover, it is easy to see that
$A\cap A^{{\perp}}=\{0\}$ since if $y\in A\cap A^{{\perp}}$ then $\langle y,y\rangle=0$, which means $y=0$.
We conclude that $X=A\oplus A^{{\perp}}$. $\square$
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