Uniform Boundedness Theorem

Also known as uniform boundedness principle and Banach–Steinhaus theorem.

Uniform Boundedness Theorem

If $\mathcal{T}$ is a set of bounded linear operators from a Banach space $X$ into a normed space $Y$ such that $\braces{\norm{Tx} : T \in \mathcal{T}}$ is a bounded set for every $x \in X$, then $\braces{\norm{T} : T \in \mathcal{T}}$ is a bounded set.

Proof: For each $n \in \NN$ the set

$$A_n = \bigcap_{T \in \mathcal{T}} \braces{x \in X : \norm{Tx} \le n}$$

is closed, since it is the intersection of the preimages of the closed interval $[0,n]$ under the continuous maps $x \mapsto \norm{Tx}$. Given any $x \in X$, the set $\braces{\norm{Tx} : T \in \mathcal{T}}$ is bounded by assumption. This means that there exists a $n \in \NN$ such that $\norm{Tx} \le n$ for all $T \in \mathcal{T}$. In other words, $x \in A_n$. Thus we have show that $\bigcup A_n = X$. XXX Apart from the trivial case $X = \emptyset$, the union $\bigcup A_n$ has nonempty interior. Now, utilizing the completeness of $X$, the Baire Category Theorem implies that there exists a $m \in \NN$ such that $A_m$ has nonempty interior. It follows that $A_m$ contains an open ball $B(y,\epsilon)$.

To show that $\braces{\norm{T} : T \in \mathcal{T}}$ is bounded, let $z \in X$ with $\norm{z} \le 1$. Then $y+\epsilon z \in B(y,\epsilon)$. Using the reverse triangle inequality and the linearity of $T$, we find

$$\epsilon \norm{Tz} \le \norm{Ty} + \norm{T(y + \epsilon z)} \le 2m.$$

This proves $\norm{T} \le 2m/\epsilon$ for all $T \in \mathcal{T}$. $\square\enspace$

---

In particular, for a sequence of operators $(T_n)$, if there are pointwise bounds $c_x$ such that

$$\norm{T_n x} \le c_x \quad \forall n \in \NN, \forall x \in X,$$

the theorem implies the existence of bound $c$ such that

$$\norm{T_n} \le c \quad \forall n \in \NN.$$

If $X$ is not complete, this may be false.

---

1. Munkres, J. (2014). Topology (2nd ed.). Pearson Education Limited.