> [!theorem] > > Let $(X, \topo)$ be a [[Normal Space|normal space]], $A \subset X$ be a [[Closed Set|closed set]] and $f \in BC(A, [a, b])$ be a [[Space of Bounded Continuous Functions|bounded continuous]] function. Then there exists a [[Continuity|continuous]] extension $F \in BC(X, [a, b])$ of $f$ to the whole space. > > *Proof*. First suppose that $f \in BC(A, [0, 1])$. > > Using the [[Method of Successive Approximations]], take $BC(A, [0, 1])$ equipped with the [[Uniform Norm|uniform norm]] as the space, with > $ > T: BC(X, [0, 1]) \to BC(A, [0, 1]) \quad f \mapsto f|_A > $ > as the continuous linear map, and $\Sigma = BC(X, [0, 1])$ as the approximating sets. > > Let $f \in BC(A, [0, 1])$ and take > $ > B = f^{-1}\paren{\norm{f} \cdot \braks{0, 1/3}} \quad C = f^{-1}\paren{\norm{f} \cdot [2/3, 1]} > $ > As $f$ is continuous, and $B, C$ are closed in $A$ under the relative topology. Since $A$ is closed, $B, C$ are closed in $X$ as well. > > By [[Urysohn's Lemma]], there exists a continuous function $\phi: X \to [0, \norm{f}/3]$ such that $\phi(B) = \bracs{0}$ and $\phi(C) = \bracs{\norm{f}/3}$. Since $\phi(x) \le \norm{f}/3$ for all $x \in X$, $\norm{\phi} \le \norm{f}$. Now, > $ > \begin{align*} > f(x) - \phi(x) &\le \norm{f}\paren{1 - \frac{1}{3}} = \frac{2}{3}\norm{f} &\forall x \in B \\ > 0 \le f(x) - \phi(x) &\le \norm{f}\paren{\frac{2}{3} - 0} \frac{2}{3}\norm{f} &\forall x \in B^c \cap A > \end{align*} > $ > with $f(x) \ge \frac{1}{3} \ge \phi(x)$ for all $x \in B^c \cap A$ and $f(x) \ge 0 = \phi(x)$ for all $x \in B$, we have $\norm{f(x) - \phi|_A(x)} \le \frac{2}{3}\norm{f}$. > > Therefore $\Sigma$ satisfies the criterion for the method of successive approximations, and for any $f \in BC(A, [0, 1])$ there exists $\seq{\phi_n} \subset BC(X, [0, 1])$ such that $\paren{\sum_{n = 1}^{\infty}\phi_n}|_A = f$. > > Now let $f \in BC(A, [a, b])$. Taking $g = \frac{f - a}{b - a}$ yields a function $g \in BC(A, [0, 1])$. Reversing the transformation on the extension yields a continuous extension of the original $f$.