# 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 $T$ of bounded linear operators from $X$ to $Y$ is

pointwise boundedif the set ${∥Tx∥:T∈T}$ is bounded for every $x∈X$,uniformly boundedif the set ${∥T∥:T∈T}$ is bounded.

Clearly, every uniformly bounded collection of operators is pointwise bounded since $∥Tx∥≤∥T∥∥x∥$. 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 $T$ is a pointwise bounded collection of bounded linear operators from $X$ to $Y$. For each $n∈N$ the set

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

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

$ϵ∥Tz∥≤∥Ty∥+∥T(y+ϵz)∥≤2m.$This proves $∥T∥≤2m/ϵ$ for all $T∈T$. $□$

In particular, for a sequence of operators $(T_{n})$, if there are pointwise bounds $c_{x}$ such that

$∥T_{n}x∥≤c_{x}∀n∈N,∀x∈X,$the theorem implies the existence of bound $c$ such that

$∥T_{n}∥≤c∀n∈N.$If $X$ is not complete, this may be false.