Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 5 / a

The sequence of functions $(f_n)$ converges uniformly to $f$ on $E$ ifThe same $N$ must work for every $x$. Equivalently, the supremum norm of $f_n-f$ tends to zero. This is stronger than pointwise convergence, where $N$ may depend on $x$.

The Uniform Cauchy criterion says that $(f_n:E\to \mathbb{R})$ converges uniformly on $E$ if and only ifFor necessity, use the triangle inequality through the uniform limit. For sufficiency, the condition makes $(f_n(x){)}_n$ a Cauchy sequence for each $x$, so completeness of the real numbers defines a pointwise limit $f(x)$. Apply the condition with $\epsilon /2$ and let $m\to \infty$ to obtain $|f_n(x)-f(x)|\le \epsilon /2<\epsilon$ uniformly in $x$.

Use uniform convergence to choose $N$ with $|f_n(x)-f_N(x)|\le 1$ for all $n\ge N$ and $x\in E$; this follows, for example, from the Uniform Cauchy criterion. Since each $f_j$ is a bounded function, put $M_j={\sup }_E|f_j|<\infty$ for $1\le j\le N$. Thenfor every $n$ and $x$. Thus $(f_n)$ is a uniformly bounded family of functions. The finite initial segment is essential: uniform convergence directly controls only a tail of a sequence.