Uniform boundedness principle

Remarks

For $k \ge 1$ let $F_k = \set{x \in X : \norm{Tx} \le k \text{ for all } T \in \mathcal{F}}$. Each $F_k$ is closed, and the hypothesis says that the $F_k$ cover $X$. Since $X$ is complete, the Baire category theorem gives some $F_k$ that contains a ball $B(x_0, r)$. Then for $\norm{x} < r$ and $T \in \mathcal{F}$, $\norm{Tx} \le \norm{T(x_0 + x)} + \norm{T x_0} \le 2k$, so $\norm{T} \le 2k/r$.

A main use is the Banach–Steinhaus theorem: if continuous linear maps $T_n \colon X \to Y$ converge at every point to $Tx$, then $T$ is linear and continuous, since $\norm{Tx} \le \sup_n \norm{T_n}\,\norm{x}$. Completeness of $X$ is necessary. Let $X$ be the space of finitely supported real sequences with the norm $\norm{x}_\infty = \sup_k \abs{x_k}$, and let $T_n(x) = n x_n$. For each $x$ the values $T_n x$ are eventually $0$, so they are bounded, but $\norm{T_n} = n$.

Going deeper

For a student

It is one of three consequences of the Baire category theorem at the core of a first course on Banach spaces. The other two are the bounded inverse theorem and the closed graph theorem. A classical application shows that there is a continuous periodic function whose Fourier series diverges at a point: the partial-sum functionals at that point have unbounded norms. It also shows that weakly convergent sequences in a normed space are bounded, by applying it to the sequence viewed in the bidual.