The Lemma to the Uniform Boundedness Principle

Lemma 1: Let $X$ be a complete metric space and let $\mathcal F$ be a collection of continuous functions on $X$. If for each $x \in X$ we have that $\displaystyle{\sup_{f \in \mathcal F} |f(x)| < \infty}$ then there is a nonempty open subset $U \subseteq X$ such that $\displaystyle{\sup_{x \in U, \: f \in \mathcal F} |f(x)| < \infty}$.

• Proof: For each $n \in \mathbb{N}$ let:

(1)

\begin{align} \quad U_n = \{ x \in X : |f(x)| \leq n, \forall f \in \mathcal F \} \end{align}

• Then $U_n$ can be alternatively represented as:

(2)

\begin{align} \quad U_n = \bigcap_{f \in \mathcal F} \{ x \in X : |f(x)| \leq n \} \end{align}

• Observe that for each $f \in \mathcal F$ we have that $\{ x \in X : |f(x)| \leq n \}$ is a closed set since $f$ is continuous. Therefore each $U_n$ is an arbitrary intersection of closed sets and is hence closed.

• Now since for each $x \in X$, $\displaystyle{\sup_{f \in \mathcal F} |f(x)| < \infty}$, there must exists an $N_x \in \mathbb{N}$ such that $x \in U_{N_x}$. Therefore:

(3)

\begin{align} \quad X = \bigcup_{n=1}^{\infty} U_n \end{align}

• Hence there exists an $n^* \in \mathbb{N}$ such that $U_{n^*}$ has nonempty interior. Let $U = \mathrm{int} (U_{n^*})$. Then for all $x \in U$ we have that $|f(x)| \leq n^*$ for all $f \in \mathcal F$, that is:

(4)

\begin{align} \quad \sup_{x \in U, \: f \in \mathcal F} |f(x)| < \infty \quad \blacksquare \end{align}