Uniform boundedness principle

If a family of bounded linear operators from a Banach space to a normed space is pointwise bounded, then their operator norms are uniformly bounded.

For operators on a complex Hilbert space, pointwise boundedness of $⟨Tx,x⟩$ over a family implies uniform boundedness of their operator norms. The complex polarization identity recovers each $⟨Tx,y⟩$ from four diagonal values. Apply the Uniform boundedness principle first to the functionals in $y$ for fixed $x$, then to the operators in $x$. On a real Hilbert space the claim fails: multiples of a nonzero skew-symmetric operator have zero diagonal forms but unbounded norms. For example, rotations through a quarter turn on ${\mathbb{R}}^2$ multiplied by integers suffice.

Let $X$ be a normed space, $Y$ a Banach space, and $A_n:X\to Y$ be bounded linear operators with $C={\sup }_n\Vert A_n\Vert <\infty$. Convergence of $A_{n}d$ for every $d$ in a dense subset $D$ implies convergence of $A_{n}x$ for every $x\in X$ to a bounded linear operator $A$, with $\Vert A\Vert \le {\mathrm{liminf}}_n\Vert A_n\Vert$. Indeed, $\Vert A_{n}x-A_{m}x\Vert \le 2C\Vert x-d\Vert +\Vert A_{n}d-A_{m}d\Vert$ makes $(A_{n}x)$ a Cauchy sequence; completeness gives its limit, and linearity and the bound pass to that limit. If $X$ is a Banach space, pointwise convergence conversely implies the uniform bound by the Uniform boundedness principle. If the limit bounded linear operator is specified in advance, completeness of $Y$ is unnecessary.

A subset of a normed space is norm bounded if it is bounded under every continuous linear functional. Apply the Uniform boundedness principle on the complete dual space to the evaluation family $f\mapsto f(x)$. Hahn-Banach theorem identifies the evaluation norm with $\Vert x\Vert$, even when the original normed space is not complete. As a consequence, two norms with exactly the same continuous dual as sets must be equivalent: apply this criterion to either norm's unit ball in the other norm.

Let $(x_n)$ lie in a normed space $X$ and suppose ${\sum }_n|f(x_n)|<\infty$ for every $f\in X^{\ast }$. Then some $C$ satisfiesApply the uniform boundedness principle on the Banach space $X^{\ast }$ to the finite signed sums $f\mapsto {\sum }_n{a}_{n}f(x_n)$ with $|a_n|\le 1$.