Uniform boundedness principle - Imported Knowlpedia Content

Statement

Let $X$ be a Banach space over $\mathbb R$ or $\mathbb C$ , let $Y$ be a normed vector space over the same field, and let $\mathcal F\subseteq B(X,Y)$ be any family of bounded linear operators. If $\mathcal F$ is pointwise bounded, meaning

\sup_{T\in\mathcal F}\lVert Tx\rVert<\infty \quad\text{for every }x\in X,

then it is uniformly bounded in operator norm:

\sup_{T\in\mathcal F}\lVert T\rVert<\infty.

The conclusion turns individual bounds into one constant valid for the whole family. No countability assumption on $\mathcal F$ is required Rudin, Chapter 2.

Proof mechanism

For each positive integer $n$ , set

E_n=\{x\in X:\sup_{T\in\mathcal F}\lVert Tx\rVert\leq n\}.

The sets $E_n$ are closed and cover $X$ . The Baire category theorem makes one $E_n$ contain a ball. Subtracting two points in that ball and using linearity bounds every $T\in\mathcal F$ on a ball about $0$ ; rescaling yields a common operator-norm bound.

Consequences and sharpness

If a sequence $T_nx$ converges for every $x\in X$ , then $(T_nx)$ is bounded for each $x$ , so $\sup_n\lVert T_n\rVert<\infty$ . This is a standard route from pointwise convergence to continuous dependence.

Completeness of the domain is essential. Let $c_{00}$ carry the supremum norm and define $T_nx=nx_n$ . Every $x\in c_{00}$ has only finitely many nonzero coordinates, so $\sup_n|T_nx|<\infty$ , but $\lVert T_n\rVert=n$ . The incomplete space $c_{00}$ is therefore a decisive near-miss.

References

John B. Conway, A Course in Functional Analysis, 2nd ed., Graduate Texts in Mathematics 96, Springer, 1990. Springer DOI record. Relevant: Chapter VI, “Linear Operators on a Banach Space.”
Walter Rudin, Functional Analysis, 2nd ed., McGraw–Hill, 1991. WorldCat record. Relevant: Chapter 2, the uniform boundedness principle.