Past exam of the functional analysis II course of the University of Oxford / 2023 / 2 / a

The Banach-Steinhaus convergence theorem can be stated as follows. Let $X$ be a normed space, $Y$ a Banach space, and $A_n\in \mathcal{B}(X,Y)$. Ifand $A_{n}d$ converges in $Y$ for every $d$ in a dense subset $D\subseteq X$, then $A_{n}x$ converges for every $x\in X$. The pointwise limit $A$ is a bounded linear operator andIf $X$ is also a Banach space, the uniform operator norm bound is necessary: pointwise convergence implies pointwise boundedness, so the Banach-Steinhaus theorem, namely the Uniform boundedness principle, supplies it. The limit is convergence in the strong operator topology; it need not be convergence in operator norm.

To see the sufficient direction, write $C={\sup }_n\Vert A_n\Vert$. For $d\in D$ the triangle inequality givesFirst approximate $x$ by $d$, then let $m,n\to \infty$. Completeness of $Y$ gives the limit. Linearity passes to limits, and $\Vert Ax\Vert ={\lim }_n\Vert A_{n}x\Vert \le ({\mathrm{liminf}}_n\Vert A_n\Vert )\Vert x\Vert$ gives the displayed bound. If a candidate bounded linear operator $A:X\to Y$ is already given and $A_{n}d\to Ad$ on $D$, the same approximation proves convergence to $Ax$ even when $Y$ is merely a normed space.

Composition with $t\mapsto t^{1+{\alpha }_n}$ preserves continuity, so $A_n$ maps the space of continuous functions on a compact space $C([0,1])$ into itself and is linear. Since $t\mapsto t^{1+{\alpha }_n}$ maps $[0,1]$ onto itself,For $0<t\le 1$, the mean value theorem applied to $r\mapsto t^{1+r}$ givesThe same bound holds at $t=0$. Each $x\in C([0,1])$ has uniform continuity by the Heine-Cantor theorem. With its modulus of continuity ${\omega }_x(\delta )={\sup }_{|s-t|\le \delta }|x(s)-x(t)|$, we obtainAlternatively, the bound proves convergence on every polynomial; the Weierstrass approximation theorem makes these a dense subset of $C([0,1])$, and the Banach-Steinhaus convergence theorem with $\Vert A_n\Vert =1$ gives the same conclusion.