Topological Complements of Normed Linear Subspaces

Recall from the Algebraic Complements of Linear Subspaces page that if $X$ is a linear space and $M \subset X$ is a linear subspace then another linear subspace $M' \subset X$ is said to be an algebraic complement of $M$ if both:

(1)

\begin{align} \quad M \cap M' = \{ 0 \} \end{align}

(2)

\begin{align} \quad X = M + M' \end{align}

We said that if a linear subspace $M$ has an algebraic complement $M'$ with finite dimensional, then $M$ is said to be finite co-dimensional. We also proved that every linear subspace $M$ of $X$ has a algebraic complement. When $X$ is a normed linear space, then we can look at algebraic complements with particular properties. One type of important algebraic complement is defined below.

Definition: Let $X$ be a normed linear space and let $M \subset X$ be a linear subspace. A Topological Complement of $M$ is an algebraic complement $M'$ that is closed.

The following theorem tells us exactly when $M \subset X$ has a topological complement when $X$ is a Banach space.

Theorem 1: Let $X$ be a Banach space. Then a linear subspace $M \subset X$ has a topological complement if and only if there exists an projection $P : X \to X$ with $P(X) = M$.

• Proof: $\Rightarrow$ Suppose that $M \subset X$ has a topological complement $M' \subset X$. Then $M \cap M' = \{ 0 \}$, $X = M \oplus M'$, and $M'$ is closed. In particular, for every $x \in X$, $x$ can be written uniquely as:

(3)

\begin{align} \quad x = m + m' \end{align}

• where $m \in M$ and $m' \in M'$. Define a function $P : X \to X$ for all $x = m + m' \in X$ by:

(4)

\begin{align} \quad P(x) =m \end{align}

• We first show that $P$ is linear. Let $x = m + m' \in X$ and $y = n + n' \in X$ (where $n \in M$, $n' \in M'$) and let $\lambda \in \mathbb{C}$. Then:

(5)

\begin{align} \quad P(x + y) = P((m + m') + (n + n')) = P((m + n) + (m' + n')) = m + n = P(x) + P(y) \end{align}

(6)

\begin{align} \quad P(\lambda x) = P(\lambda (m + m')) = P(\lambda m + \lambda m') = \lambda m = \lambda P(x) \end{align}

• So indeed, $P$ is linear. Furthermore, if $x = m + m' \in X$ then:

(7)

\begin{align} \quad P^2(x) = P(P(x)) = P(P(m + m')) = P(m + 0) = m = P(x) \end{align}

• Therefore $P$ is idempotent. Lastly, we show that $P(X) = M$. Clearly $P(X) \subseteq M$. Now let $m \in M$ . Then take $x = m \in X$. So $P(x) \in P(X)$. But $P(x) = m$, so $m \in P(X)$. Thus $P(X) \supseteq M$. So $P(X) = M$.

• $\Leftarrow$ Suppose that there exists an idempotent linear operator $P : X \to X$ with $P(X) = M$. Let $M' = \ker P$. Then by the lemma on the Projection/Idempotent Linear Operators page we have that:

(8)

\begin{align} \quad M' = \ker P = (I - P)(X) \end{align}

• In particular, $M'$ is closed since $M' = \ker P = P^{-1}(\{ 0 \})$ and $P$ is a continuous map. Furthermore:

(9)

\begin{align} \quad X = P(X) \oplus (I - P)(X) = M \oplus M' \end{align}

• So $M'$ is a topological complement of $M$. $\blacksquare$