Corollaries to the Uniform Boundedness Principle

# Corollaries to the Uniform Boundedness Principle

Recall from The Uniform Boundedness Principle page that if $X$ is a Banach space and $Y$ is a normed linear space where $\mathcal F$ is a collection of bounded linear operators from $X$ to $Y$ such that for every $x \in X$ we have that $\sup_{T \in \mathcal F} \| T(x) \| < \infty$ then:

(1)
\begin{align} \quad \sup_{T \in \mathcal F} \| T \| < \infty \end{align}

We will now look at some corollaries to the uniform boundedness principle.

 Corollary 1: Let $X$ be a Banach space and let $A \subseteq X^*$. Then $A$ is bounded in $X^*$ if and only if for every $x \in X$ we have that $\sup_{f \in A} |f(x)| < \infty$.
• Proof: Recall that $X^* = \mathcal B(X, \mathbb{R})$.
• $\Leftarrow$ Suppose that $A$ is bounded. Then there exists an $M > 0$ such that for all $f \in A$ we have that $\| f \| < M$. Let $x \in X$. Then:
(2)
\begin{align} \quad \sup_{f \in A} |f(x)| \leq \sup_{f \in A} \| f \| \| x \| = \| x \| \sup_{f \in A} \| f \| \leq \| x \| M < \infty \end{align}
• $\Rightarrow$ Since $X$ is a Banach space and $A$ is a collection of bounded linear operators from $X$ to $\mathbb{R}$, and since for every $x \in X$ we are given that $\sup_{f \in A} |f(x)| < \infty$, by the principle of uniform boundedness we have that:
(3)
\begin{align} \quad \sup_{f \in A} \| f \| < \infty \end{align}
• So let $\sup_{f \in A} \| f \| = M$. Then for all $f \in A$ we have that $\| f \| \leq M$ which shows that $A$ is bounded. $\blacksquare$
 Corollary 2: Let $X$ and $Y$ be Banach spaces. If $(T_n)$ is a sequence of bounded linear operators from $X$ to $Y$ such that for every $x \in X$, $\displaystyle{\lim_{n \to \infty} T_n(x)}$ exists, then $\displaystyle{T(x) = \lim_{n \to \infty} T_n(x)}$ is a bounded linear operator from $X$ to $Y$.
• Proof: Let $\mathcal F = \{ T_n \}$. For each $x \in X$ we have that $\displaystyle{\lim_{n \to \infty} T_n(x) = y_x}$ for some $y_x \in Y$. Therefore $\displaystyle{\lim_{n \to \infty} \| T_n(x) - y_x \| = 0}$ and by the triangle inequality, $\lim_{n \to \infty} \| T_n(x) \| = \| y_x \|$. So for each $x \in X$ we see that the sequence $(T_n(x))$ is bounded, and so for each $x \in X$:
(4)
\begin{align} \quad \sup_{n \geq 1} \| T_n(x) \| < \infty \end{align}
• Since each $T_n$ is a bounded linear operator and $X$ is a Banach space, by the uniform boundedness principle:
(5)
\begin{align} \quad \sup_{n \geq 1} \| T_n \| < \infty \end{align}
• In other words, there exists an $M \in \mathbb{R}$, $M \geq 0$ such that:
(6)
\begin{align} \quad \sup_{n \geq 1} \| T_n \| = M \end{align}
• And so for every $x \in X$ we have that $\| T_n(x) \| \leq M \| x \|$. Taking the limit as $n \to \infty$ gives us:
(7)
\begin{align} \quad \| T(x) \| = \left \| \lim_{n \to \infty} T_n(x) \right \| = \lim_{n \to \infty} \| T_n(x) \| \leq M \| x \| \end{align}
• So $T$ is bounded. $\blacksquare$
 Theorem 3 (The Banach-Steinhaus Theorem): Let $X$ and $Y$ be Banach spaces and let $\{ T_n \}$ be a sequence of bounded linear operators from $X$ to [[ $Y$]] such that for every $x \in X$ there exists a $y \in Y$ for which $\lim_{n \to \infty} \| T_n(x) - y \| = 0$. Then there exists a a bounded linear operator $T : X \to Y$ such that for every $x \in X$ we have that $\lim_{n \to \infty} \| T_n(x) - T(x) \| = 0$ and $\sup \| T_n \| < \infty$.
• Proof: Let $T : X \to Y$ be defined for each $x \in X$ by:
(8)
\begin{align} \quad T(x) = \lim_{n \to \infty} T_n(x) = y \end{align}
• Note that $T$ is is well-defined and furthermore, $T$ is linear since limits are linear. We will now show that $T$ is bounded.
• Let $\mathcal F = \{ T_n : n \in \mathbb{N} \}$. Then $\mathcal F$ is a collection of bounded linear operators from $X$ to $Y$. For each $x \in X$ since $T_n(x)$ converges to $y$ we must have that $\{ \| T_n(x) \| : n \in \mathbb{N} \}$ is bounded, that is, for each $x \in X$ we have that:
(9)
\begin{align} \quad \sup_{n \geq 1} \| T_n(x) \| < \infty \end{align}
• So by the uniform boundedness principle we have that:
(10)
\begin{align} \quad \sup_{n \geq 1} \| T_n \| < \infty \end{align}
• Let $M = \sup_{n \geq 1} \| T_n \|$. Then for all $n \in \mathbb{N}$ we have that $\| T_n \| \leq M$. Suppose that $x \in X$ is such that $\| x \| \leq 1$. Then for all $n \in \mathbb{N}$ we have that:
(11)
\begin{align} \quad \| T(x) \| = \| T(x) - T_n(x) + T_n(x) \| \leq \| T(x) - T_n(x) \| + \| T_n(x) \| \leq \|T(x) - T_n(x) \| + M \leq \| T - T_n(x) \| \| x \| + M \leq \| T - T_n \| + M \end{align}
• Taking the limit as $n \to \infty$ gives us that:
(12)
\begin{align} \quad \| T(x) \| \leq M \end{align}
• Hence:
(13)
\begin{align} \quad \| T \| = \sup_{x \in X, \| x \| \leq 1} \| T(x) \| \leq M \end{align}
• Which implies that $T$ is a bounded linear operator. Lastly, for all $x \in X$ we have that $\lim_{n \to \infty} \| T_n(x) - T(x) \| = \lim_{n \to \infty} \| T_n(x) - y \| = 0$ and by the principle of uniform boundedness we also have that $\sup \| T_n \| < \infty$. $\blacksquare$