Example

— *1) Let $\mu$ be the Lebesgue measure on $\mathbf{R}$, and let E be the semi-normed space $\mathscr{L}^p(\mu)$ ($1 \leq p < \infty$) (INT, IV). For every numerical function $f$ and every real number $h$, let $f_h$ be the function $x \mapsto f(x - h)$. Clearly the mapping $f \mapsto f_h$ defines a linear isometry from E onto itself. If $f$ is continuous and has compact support, then $f_h$ converges to $f$ uniformly, hence also in the mean of order $p$, as $h$ tends to 0. Since the set $\mathscr{K}(\mathbf{R})$ of all continuous functions with compact support is dense in E, and the set of linear isometries of E is equicontinuous, it follows from prop. 5 that for every $f \in E$, $f_h$ converges in the mean of order $p$ to $f$ as $h$ tends to 0.

For $p = 1$, consider the Fourier transform, which associates to each $f \in \mathscr{L}^1(\mu)$ the function $\hat{f}$ on $\mathbf{R}$ defined by

The set of linear forms $f \mapsto \hat{f}(y)$ is an equicontinuous subset of the dual of $\mathscr{L}^1(\mu)$.

On the other hand, we know that the set T of all characteristic functions of closed bounded intervals is a total subset of $\mathscr{L}^1(\mu)$; and we verify easily that for all $f \in T$, the Fourier transform $\hat{f}$ is a continuous function tending to zero at infinity. We deduce that this is true for all $f \in \mathscr{L}^1(\mu)$ ("Riemann-Lebesgue theorem").

The relation $\sup_{y \in \mathbf{R}} |\hat{f}(y)| \leq \|f\|_1$ shows that the map $f \mapsto \hat{f}$ is a continuous map from $\mathscr{L}^1(\mu)$ into the space $\mathscr{B}(\mathbf{R})$ of all bounded functions on $\mathbf{R}$, with the structure of uniform convergence. Since $\hat{f}$ is continuous for all $f \in T$, it follows that $\hat{f}$ is continuous for all $f \in L^1(\mu)$. The fact that $\hat{f}$ tends to zero at infinity follows from the fact that the subspace $C_0(\mathbf{R})$ of all continuous functions tending to zero at infinity is closed in $\mathscr{B}(\mathbf{R})$.

2) Let E be the space of all continuous numerical functions on $\mathbf{R}$ endowed with the topology of compact convergence. Let K be a compact subset of $\mathbf{R}$ and let $(\mu_n)$ be a sequence of measures on $\mathbf{R}$ with support in K. Suppose $\|\mu_n\| \leq 1$ for all n. The set of the $\mu_n$ is then an equicontinuous subset of E'. Therefore, if for every function $f \in E$, we have $\lim_{n \to \infty} \mu_n(f) = 0$, the sequence of functions $x \mapsto \int e^{itx} d\mu_n(t)$ converges to 0, uniformly on every compact subset of $\mathbf{R}$ (since the set of functions $t \mapsto e^{itx}$, as x ranges over a compact subset of $\mathbf{R}$, is compact in E). \*