Uniform Boundedness Theorem

Also known as Uniform Boundedness Principle and Banach–Steinhaus Theorem.

Definition (Pointwise and Uniform Boundedness)

Let $X$, $Y$ be normed spaces. We say that a collection $\mathcal{T}$ of bounded linear operators from $X$ to $Y$ is

• pointwise bounded if the set $\{\Vert Tx\Vert :T\in \mathcal{T}\}$ is bounded for every $x\in X$,
• uniformly bounded if the set $\{\Vert T\Vert :T\in \mathcal{T}\}$ is bounded.

Clearly, every uniformly bounded collection of operators is pointwise bounded since $\Vert Tx\Vert \le \Vert T\Vert \Vert x\Vert$. The converse is true, if $X$ is complete:

Uniform Boundedness Theorem

If a collection of bounded linear operators from a Banach space into a normed space is pointwise bounded, then it is uniformly bounded.

Proof   Suppose $X$ is a Banach space, $Y$ is a normed space and $\mathcal{T}$ is a pointwise bounded collection of bounded linear operators from $X$ to $Y$. For each $n\in \mathbb{N}$ the set

$$A_n=\bigcap _{T\in \mathcal{T}}\{x\in X:\Vert Tx\Vert \le n\}$$

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

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

$$ϵ\Vert Tz\Vert \le \Vert Ty\Vert +\Vert T(y+ϵz)\Vert \le 2m.$$

This proves $\Vert T\Vert \le 2m/ϵ$ for all $T\in \mathcal{T}$. $\square$

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

$$\Vert T_{n}x\Vert \le c_x\forall n\in \mathbb{N},\forall x\in X,$$

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

$$\Vert T_n\Vert \le c\forall n\in \mathbb{N}.$$

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