Definition 5.5.1.
The orthogonal complement of a subset \(E\) of a Hilbert space \(\hilbert\) is the set
\begin{equation*}
E^\perp = \{x \in \hilbert: \ip{x}{y} = 0 \text{ for all } y \in E\}.
\end{equation*}
Section 5.5 Orthogonal complements
One of the most useful properties of orthogonality in Euclidean space is to decompose the space into orthogonal subspaces - this allows vectors to be expressed uniquely as sums of components lying in each subspace, for example. (You will have done this implicitly when studying projection onto subspaces, for example.) The same geometric properties hold in Hilbert space.
Checkpoint 5.5.2.
Let
\begin{equation*}
E = \{f \in L^2(0,1): f(t) = 0 \text{ for } t \in (0, \frac{1}{2})\}.
\end{equation*}
Describe \(E^\perp\) in \(L^2(0,1)\text{.}\)
It is a useful fact that orthogonal complements are always closed subspaces relative to the larger Hilbert space.
Theorem 5.5.3.
For any set \(E \subset \hilbert\text{,}\) the orthogonal complement \(E^\perp\) is a closed linear subspace of \(\hilbert\text{.}\)
Proof.
Elements in an orthgonal complement of a subspace are precisely those that satisfy a sort of “least distance” property - that is, perturbing \(x\) by any vector in \(E\) doesn’t get you any close to \(E\text{.}\) This leads to a nice norm characterization of the orthogonal complement.
Lemma 5.5.4.
Let \(E\) be a linear subspace of an inner produt space \(\hilbert\) and let \(x\) be a vector in \(\hilbert\text{.}\) Then \(x \in E^\perp\) if and only if
\begin{equation}
\norm{x - y} \geq \norm{x} \text{ for all } y \in E.\tag{5.5.1}
\end{equation}
Proof.
(\(\Rightarrow\)): If \(x \in E^\perp\text{,}\) then for any \(y \in E\text{,}\) \(x\) and \(y\) are orthogonal. Then Theorem 5.1.4 gives
\begin{equation*}
\norm{x - y}^2 = \norm{x}^2 + \norm{y}^2 \geq \norm{x}^2.
\end{equation*}
(\(\Leftarrow)\text{:}\) Now suppose that the inequality in (5.5.1) holds. Pick an arbitrary \(y \in E\) and a scalar \(c \in \C\text{.}\) As \(E\) is a linear subspace, \(cy \in E\text{,}\) and so
\begin{equation*}
\norm{x - cy}^2 \geq \norm{x}^2.
\end{equation*}
Now we leverage inner product geometry. Notice that
\begin{equation*}
- 2\RE \cc{c} \ip{x}{y} + \abs{c}^2 \norm{y}^2 \geq 0.
\end{equation*}
Now, we’ll drop the real part of the inner product. Let \(z\) be a unimodular constant so that \(\cc{z}\ip{x}{y} = \abs{\ip{x}{y}}\) (if you don’t see why such a constant should exist, you should think about it). Then the inequality above holds in particular for vectors of the form \(y = tz\) where \(t \in \R^+\text{,}\) which gives
\begin{equation*}
- 2 t\RE \ip{x}{y} + t^2 \norm{y}^2 \geq 0.
\end{equation*}
We can rearrange this into
\begin{equation*}
\abs{\ip{x}{y}} \leq \frac{1}{2} t \norm{y}^2,
\end{equation*}
for all \(t \in \R^+\text{.}\) Letting \(t \to 0\text{,}\) we conclude that \(x \perp y\) for all \(y \in E\text{,}\) and so \(x \in E^\perp\text{.}\)
Now we can produce the Hilbert space analogue of decomposing a vector into orthogonal components with repect to orthogonal subspaces.
Theorem 5.5.5.
Let \(E\) be a closed linear subspace of a Hilbert space \(\hilbert\text{,}\) and suppose that \(x \in \hilbert\text{.}\) Then there exists \(y \in E\) and \(z \in E^\perp\) so that \(x = y + z\text{.}\)
Proof.
The main idea of the proof is to take \(y\) to be the closest point to \(x\) in \(E\text{.}\) That is, following Theorem 4.3.2, \(y\) is the point in \(E\) so that \(\norm{x - y} \leq \norm{x - v}\) for all \(v \in E\text{.}\) Now define \(z = x - y\) so that \(x = y + z\text{.}\) Now, for any \(v \in E\text{,}\) we know that \(y + v \in E\) and so since \(y\) is the closest point to \(x\text{,}\) we have
\begin{equation*}
\norm{z} = \norm{x - y} \leq \norm{x - (y + v)}.
\end{equation*}
That is, for all \(v \in E\text{,}\) we have
\begin{equation*}
\norm {z} \leq \norm {z - v}.
\end{equation*}
By Lemma 5.5.4, we conclude that \(z \in E^\perp\text{.}\)
Corollary 5.5.6.
If \(E\) is a closed linear subspace of a Hilbert space \(\hilbert\text{,}\) then \((E^\perp)^\perp = E\text{.}\)
Proof.
Every element of \(E\) is orthogonal to \(E^\perp\) by definition, and so \(E \subseteq E^\perp\text{.}\) On the other hand, let \(x \in (E^\perp)^\perp\text{.}\) Then by Theorem 5.5.5, \(x = y + z\) where \(y \in E\) and \(z \in E^\perp\text{.}\) Because \(x \in (E^\perp)^\perp\text{,}\) we have \(\ip{x}{z} = 0\text{.}\) But then
\begin{equation*}
0 = \ip{x}{z} = \ip{y + z}{z} = \ip{y}{z} + \ip{z}{z} = \norm{z}^2,
\end{equation*}
and so \(z = 0\text{.}\) This means that \(x = y\) and thus \(x \in E\) and \((E^\perp)^\perp \subseteq E\text{.}\) The two-way inclusion proves the claim.
Definition 5.5.7.
Let \(E, F\) be subspaces of a vector space \(V\text{.}\) \(V\) is the direct sum of \(E\) and \(F\text{,}\) denoted \(V = E \oplus F\text{,}\) if \(M \cap N = \{0\}\) and every vector in \(V\) can be written as a sum of a vector in \(E\) and a vector in \(F\text{.}\) If \(V\) is an inner product space and \(E \perp F\) (in the sense that \(\ip{x}{y} = 0\) for \(x \in E, y \in F\) then \(V\) is the orthogonal direct sum of \(E\) and \(F\text{.}\)
It is immediate that for any closed linear subspace \(E\) of a Hilbert space \(\hilbert\) that \(\hilbert\) is the orthogonal direct sum \(E \oplus E^\perp\text{.}\)
Checkpoint 5.5.8.
Show that \(L^2(-1,1)\) can be orthongonally decomposed into the direct sum of its subspaces of even and odd functions.
