Theorem 4.10.3 (Tietze Extension Theorem). Let $X$ be a normal space, $A \subset X$ be closed, $U \in \cn(A)$, and $f \in BC(A; \real)$, then there exists $F \in BC(X; \real)$ such that $F|_{A} = f$ and $\norm{F}_{u} = \norm{f}_{u}$.
Proof. Let
then $R \in L(BC(X; \real); BC(A; \real))$.
For any $g \in C(A; [0, 1])$, let
then $B, C \subset A$ are closed with $B \cap C = \emptyset$. Since $A$ is closed, $B, C \subset X$ are closed. By Urysohn’s lemma, there exists $h \in C(X; [0, 1/3])$ such that $h|_{C} = 1/3$ and $h|_{B} = 0$. Thus $g - h|_{A} \in C(A; [0, 2/3])$.
By linearity, this implies that for any $g \in BC(A; \real)$, there exists $h \in BC(A; \real)$ such that $\norm{h}_{u} \le \norm{g}_{u}/3$ and $\norm{g - h|_A}_{u} \le 2\norm{g}_{u}/3$.
Since $\real$ is complete, so is $BC(X; \real)$ by Proposition 6.1.5. Using successive approximations, for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_{A} = f$ and