Complements of Closed Subspaces of Hilbert Spaces

# Complements of Closed Subspaces of Hilbert Spaces

Recall from the Orthogonal Sets in an Inner Product Space page that if $H$ is an inner product space and $S \subseteq H$ then the orthogonal set of $S$ is defined as:

(1)\begin{align} \quad S^{\perp} = \{ x \in H : x \perp S \} \end{align}

Also recall from The Best Approximation Theorem for Hilbert Spaces page that if $H$ is a Hilbert space and $K \subset H$ is a subset then if $K$ is nonempty, closed, and convex and $h_0 \in H \setminus K$ then there exists a unique vector $k_* \in K$ such that:

(2)\begin{align} \quad \| h_0 - k_* \| = \inf_{k \in K} \| h_0 - k \| \end{align}

That is, there exists a unique vector $k_*$ in $K$ that is closest to $h_0$.

We now use this theorem to proved that if $H$ is a Hilbert space then $H$ is the direct sum of any closed subspace $M$ and $M^{\perp}$.

Proposition 1: Let $H$ be a Hilbert space and let $M \subset H$ be a subspace. If $M$ is closed then $H = M \oplus M^{\perp}$. |

**Proof:**Let $x \in M \cap M^{\perp}$. Since $x \in M^{\perp}$ we have that $\langle x, y \rangle = 0$ for every $y \in M$. In particular, $x \in M$ and so $\langle x, x \rangle = 0$. But this happens if and only if $x = 0$. So:

\begin{align} \quad M \cap M^{\perp} = \{ 0 \} \end{align}

- Let $h_0 \in H \setminus M$. Since $M$ is a subspace of $H$, $M$ is nonempty and convex, and $M$ is given to be closed. So by the theorem of best approximation there exists a unique vector $m_* \in M$ such that:

\begin{align} \quad \| h_0 - m_* \| = \mathrm{inf}_{m \in M} \| h_0 - m \| \end{align}

- Let $x \in M$. Since $M$ is a subspace of $H$, for every $t \in \mathbb{R}$ we have that $m_* - tx \in M$. Therefore, from the inequality above we have that:

\begin{align} \quad \| h_0 - m_* \| \leq \| h_0 - (m_* - tx) \| = \| (h_0 - m_*) - tx \| \end{align}

- We square both sides of the inequality above to get:

\begin{align} \quad \| h_0 - m_* \|^2 & \leq \| (h_0 - m_*) - tx \|^2 \\ & \leq \langle (h_0 - m_*) - tx, (h_0 - m_*) - tx \rangle \\ & \leq \| h_0 - m_* \|^2 + 2t \mathrm{Re} \langle x, h_0 - m_* \rangle + t^2 \| x \|^2 \end{align}

- Therefore, for every $t \in \mathbb{R}$

\begin{align} \quad 0 \leq 2t \mathrm{Re} \langle x, h_0 - m_* \rangle + t^2 \| x \|^2 \end{align}

- Observe that $t = 0$ satisfies the inequality above, and so the polynomial $2t \mathrm{Re} \langle x, h_0 - m_* \rangle + t^2 \| x \|^2$ has a root of $t = 0$ of multiplicity $2$. Therefore:

\begin{align} \quad \langle x, h_0 - m_* \rangle = 0 \end{align}

- Since this holds for every $x \in M$ we have that $h_0 - m_* \in M^{\perp}$. Therefore we have that:

\begin{align} \quad h_0 = \underbrace{m_*}_{\in M} + \underbrace{h_0 - m_*}_{\in \: M^{\perp}} \end{align}

- Hence $H = M + M^{\perp}$, and so:

\begin{align} \quad H = M \oplus M^{\perp} \quad \blacksquare \end{align}