Fin.-Dim. Sub. of Normed Lin. Sps. have Topological Complements

# Finite-Dimensional Subspaces of Normed Linear Spaces have Topological Complements

Theorem 1: Let $X$ be a normed linear space and let $Y \subseteq X$ be a subspace. If $Y$ is finite-dimensional then $Y$ has a topological complement. |

**Proof:**Let $Y \subseteq X$ be a finite-dimensional subspace. Then for some $n \in \mathbb{N}$ we have that:

\begin{align} \quad \dim (Y) = n \end{align}

- Let $\mathcal B = \{ e_1, e_2, ..., e_n \}$ be a basis for $Y$. Then every $y \in Y$ can be uniquely expressed as a linear combination of the form:

\begin{align} \quad y = a_1e_1 + a_2e_2 + ... + a_ne_n \end{align}

- Where $a_1, a_2, ..., a_n \in \mathbb{C}$. For each $k \in \{ 1, 2, ..., n \}$ define a function $\psi_k : Y \to \mathbb{C}$ by:

\begin{align} \quad \psi_k(y) = \psi_k(a_1e_1 + a_2e_2 + ... + a_ne_n) = a_k \end{align}

- It is easy to verify that each $\psi_k$ is a linear functional. Furthermore, each $\psi_k$ is continuous since $Y$ is finite-dimensional. Now by the theorem on the Extensions of Linear Functionals with Equal Norms page, for each $\psi_k : Y \to \mathbb{C}$ there exists a linear functional extension $\Psi_k : X \to \mathbb{C}$ such that for each $k \in \{ 1, 2, ..., n \}$:

\begin{align} \quad \Psi_k(y) &= \psi_k(y), \quad \forall y \in Y \\ \quad \| \Psi_k \| &= \| \psi_k \| \end{align}

- Let $Y' \subseteq X$ be defined as:

\begin{align} \quad Y' = \bigcap_{k=1}^{n} \ker \Psi_k \end{align}

- We claim that $Y'$ is a topological complement of $Y$.

- Since $\ker \Psi_k$ is closed for each $k$ we have that $Y'$ is closed. Now let $y \in Y \cap Y'$. Since $y \in Y$ we have that $y = a_1e_1 + a_2e_2 + ... + a_ne_n$ and since $y \in Y'$ we have that $\Psi_k(y) = 0$ for all $k \in \{ 1, 2, ..., n \}$. Therefore:

\begin{align} \quad a_k = \Psi_k(y) = 0 \\ \end{align}

- Hence $y = 0$. Therefore:

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

- Now let $x \in X$. Observe that for each $k$ we have that:

\begin{align} \quad \Psi_k(x) = \Psi_k \left ( \sum_{j=1}^{n} \Psi_j(x)e_k \right ) \end{align}

- Therefore we have that:

\begin{align} \quad x - \sum_{j=1}^{n} \Psi_j(x)e_j \in Y' \end{align}

- So for each $x \in X$ we can write:

\begin{align} \quad x = \underbrace{\sum_{k=1}^{n} \Psi_k(x)e_k}_{\in Y} + \underbrace{\left [x - \sum_{j=1}^{n} \Psi_k(x) \in Y' \right]}_{\in Y'} \end{align}

- So $X = Y + Y'$. Therefore:

\begin{align} \quad X = Y \oplus Y' \end{align}

- So $Y'$ is a topological complement of $Y$. $\blacksquare$