The Uniform Boundedness Principle
 Table of Contents

# The Uniform Boundedness Principle

Recall from The Lemma to the Uniform Boundedness Principle page that if $X$ is a complete metric space and $\mathcal F$ is a collection of continuous functions on $X$ then if for each $x \in X$, $\displaystyle{\sup_{f \in \mathcal F} |f(x)| < \infty}$ then there exists a nonempty open set $U \subseteq X$ such that:

(1)
\begin{align} \quad \sup_{x \in U, \: f \in \mathcal F} |f(x)| < \infty \end{align}

We will use this result to prove the uniform boundedness principle.

 Theorem 1 (The Uniform Boundedness Principle): Let $X$ be a Banach space and let $Y$ be a normed linear space. Let $\mathcal F$ be a collection of bounded linear operators from $X$ to $Y$. If for every $x \in X$ we have that $\displaystyle{\sup_{T \in \mathcal F} \| T(x) \| < \infty}$ then $\displaystyle{\sup_{T \in \mathcal F} \| T \| < \infty}$.
• Proof: Let $\mathcal F$ be a collection of bounded linear operators from $X$ to $Y$. For each $T \in \mathcal F$ define the functions $f_T : X \to \mathbb{R}$ for each $x \in X$ by:
(2)
\begin{align} \quad f_T(x) = \| T(x) \| \end{align}
• Denote this set of functions by $\mathcal F^*$. Observe that each $f_T$ is a continuous function since norms are continuous and each $T$ is continuous. Furthermore, for each $x \in X$, by assumption we have that:
(3)
\begin{align} \quad \sup_{T \in \mathcal F} |f_T(x)| = \sup_{T \in \mathcal F} \| T(x) \| < \infty \quad (*) \end{align}
• By the lemma to the uniform boundedness principle, since $X$ is a Banach space (and hence complete) and for every $x \in X$, $(*)$ holds, we have that there is a nonempty open set $U \subseteq X$ such that $\displaystyle{\sup_{x \in U, \: f_T \in \mathcal F^*} |f_T(x)| \leq M < \infty}$. Since $U$ is an open set, it must contain a closed ball. So there exists an $r > 0$ and an $x_0 \in X$ such that if $\| x - x_0 \| \leq r$ and $T \in \mathcal F$ then:
(4)
\begin{align} \quad \| T(x) \| \leq M \end{align}
• Now let $x \in X$ be such that $\| x \| = 1$. Let:
(5)
\begin{align} \quad x' = x_0 + rx \end{align}
• Then we have that:
(6)
\begin{align} \quad \| x' - x_0 \| = \| x_0 + rx - x_0 \| = \| rx \| = r \| x \| \leq r \cdot 1 = r \end{align}
• Therefore, for every $T \in \mathcal F$ we have that:
(7)
\begin{align} \quad \| T(x') \| \leq M \end{align}
• Hence we have that:
(8)
\begin{align} \quad \| T(x) \| &= \biggr \| T \left ( \frac{1}{r} [x' - x_0] \right ) \biggr \| \\ &= \frac{1}{r} \| T(x' - x_0) \| \\ &\leq \frac{1}{r} [ \| T(x') \| + \| T(x_0) \| ] \\ &\leq \frac{1}{r} [ M + \| T(x_0) \|] \\ &\leq \frac{1}{r} \left [ M + \sup_{T \in \mathcal F} \|T(x_0) \| \right ] \\ & < \infty \end{align}
• Therefore we have that:
(9)
\begin{align} \quad \sup_{T \in \mathcal F} \| T \| < \infty \quad \blacksquare \end{align}
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