diff --git a/src/fa/norm/index.tex b/src/fa/norm/index.tex
index a7a7b70..d7a2680 100644
--- a/src/fa/norm/index.tex
+++ b/src/fa/norm/index.tex
@@ -9,3 +9,4 @@
 \input{./hilbert.tex}
 \input{./compact.tex}
 \input{./ap.tex}
+\input{./schauder.tex}
diff --git a/src/fa/norm/schauder.tex b/src/fa/norm/schauder.tex
new file mode 100644
index 0000000..f6be58b
--- /dev/null
+++ b/src/fa/norm/schauder.tex
@@ -0,0 +1,48 @@
+\section{Schauder Bases}
+\label{section:schauder-bases}
+
+\begin{definition}[Schauder Basis]
+\label{definition:schauder-basis}
+    Let $E$ be a separable Banach space over $K \in \RC$ and $\seq{x_n} \subset E$, then $\seq{x_n}$ is a \textbf{Schauder basis} of $E$ if for each $x \in E$, there exists a unique $\seq{\lambda_n(x)} \in K^\natp$ such that
+    \[
+    x = \limv{N}\sum_{n = 1}^N \lambda_n(x) x_n
+    \]
+
+    The sequence of mappings $\seq{\lambda_n} \subset K^{E}$ is the \textbf{coefficient forms} of $\seq{x_n}$, and the Schauder basis $\seq{x_n}$ is \textbf{normalised} if $\norm{x_n}_E = 1$ for all $n \in \natp$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:schauder-basis-functional}
+    Let $E$ be a separable Banach space over $K \in \RC$, $\seq{x_n} \subset E$ be a normalised Schauder basis, and $\seq{\lambda_n} \subset K^{E}$ be its coefficient forms, then:
+    \begin{enumerate}
+        \item $\seq{\lambda_n} \subset E^*$ is equicontinuous. 
+        \item For each $N \in \natp$ and $x \in E$, let $P_Nx = \sum_{n = 1}^N \dpn{x, \lambda_n}{E}x_n$, then $P_N \to \text{Id}$ uniformly on compact sets as $N \to \infty$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[III.9.6]{SchaeferWolff}}}. ]
+    By uniqueness of the basis decomposition, $\seq{\lambda_n} \subset \hom(E; K)$. For each $x \in E$, let
+    \[
+    \norm{x}_{E'} = \sup_{N \in \natp} \norm{\sum_{n = 1}^N \lambda_n(x)x_n}_E
+    \]
+
+    then $\norm{x}_{E} \le \norm{x}_{E'}$, and $E$ is complete with respect to $\norm{\cdot}_{E'}$. By the \hyperref[Open Mapping Theorem]{theorem:open-mapping}, $\norm{\cdot}_E$ is equivalent to $\norm{\cdot}_{E'}$, and there exists $C \ge 0$ such that $\norm{x}_{E'} \le C\norm{x}_{E}$ for all $x \in E$. 
+
+    (1): For each $N \in \natp$ and $x \in E$, since $\seq{x_n}$ is normalised, 
+    \begin{align*}
+        |\lambda_N(x)| &= \norm{\lambda_N(x)x_N}_E = \norm{\sum_{n = 1}^{N}\lambda_n(x)x_n - \sum_{n = 1}^{N-1}\lambda_n(x)x_n }_E \\
+        &\le 2\norm{x}_{E'} \le 2C\norm{x}_{E}
+    \end{align*}
+
+    Hence $\seq{\lambda_n} \subset E^*$ with $\sup_{n \in \natp}\norm{\lambda_n}_{E^*} \le 2C$. 
+    
+    (2): Since $\sup_{N \in \natp}\norm{P_N}_{L(E; E)} \le C$, $\seq{P_N}$ is equicontinuous. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $P_N \to \text{Id}$ uniformly on compact sets as $N \to \infty$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:separable-schauder-ap}
+    Let $E$ be a separable Banach space over $K \in \RC$ with a Schauder basis, then $E$ enjoys the approximation property. 
+\end{corollary}
+\begin{proof}
+    By \autoref{proposition:schauder-basis-functional}, there exists $\seq{P_N} \subset L(E; E)$ such that $P_N \to \text{Id}$ uniformly on compact sets as $N \to \infty$. 
+\end{proof}
+