---
title: Uniform Boundedness Theorem
parent: The Fundamental Four
grand_parent: Functional Analysis Basics
nav_order: 2
---

# {{ page.title }}

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
{: .mb-0 }
- *pointwise bounded* if the set $\braces{\norm{Tx} : T \in \mathcal{T}}$ is bounded for every $x \in X$,
- *uniformly bounded* if the set $\braces{\norm{T} : T \in \mathcal{T}}$ is bounded.
{% enddefinition %}

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

{% theorem * 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.
{% endtheorem %}

{% 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 $n \in \NN$ 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 shown that $\bigcup A_n = X$.
In particular, $\bigcup A_n$ has nonempty interior.
Now, utilizing the completeness of $X$, the
[Baire Category Theorem]({% link pages/general-topology/baire-spaces.md %})
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 $T \in \mathcal{T}$.
{% endproof %}

---

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.

TODO:
- strong operator convergence
- Kreyszig 4.9-5
- Haase 15.6