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 $A^{⟂}$, is a topological complement of $A$. That means $A^{⟂}$ is closed and

$$X=A\oplus A^{⟂}.$$

Proof :

• $A^{⟂}$ is closed :

  This follows easily from the continuity of the inner product. If a sequence $(x_n)$ of elements in $A^{⟂}$ converges to an element $x_0\in X$, then

  $$⟨x_0,a⟩=⟨\underset{n\to \mathrm{\infty }}{lim}{x}_n,a⟩=\underset{n\to \mathrm{\infty }}{lim}⟨x_n,a⟩=0\text{for every}a\in A$$

  which implies that $x_0\in A^{⟂}$.

• $X=A\oplus A^{⟂}$ :

  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 $x-a_0\in A^{⟂}$ (see this entry (http://planetmath.org/BestApproximationInInnerProductSpaces)).

  This allows one to write $x$ as a sum of elements in $A$ and $A^{⟂}$

  $$x=a_0+(x-a_0)$$

  which proves that

  $$X=A+A^{⟂}.$$

  Moreover, it is easy to see that

  $$A\cap A^{⟂}=\{0\}$$

  since if $y\in A\cap A^{⟂}$ then $⟨y,y⟩=0$, which means $y=0$.

  We conclude that $X=A\oplus A^{⟂}$. $\mathrm{\square }$