Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 5 / b

Let $S_N={\sum }_{n=1}^N{f}_n$ and let $s$ be its uniform limit. Choose a finite $B\ge 0$ with $|g(x)|\le B$ for all $x\in E$. The partial sums of the multiplied series are $g{S}_N$, andFor any $\epsilon >0$, uniform convergence supplies $N_0$ making $|S_N(x)-s(x)|<\epsilon /(B+1)$ for all $N\ge N_0$ and $x\in E$. The displayed estimate then gives an error less than $\epsilon$. Thus

Choose the unbounded function $\boxed{g(x)=1/x}$ on $E=(0,1/2]$. It is unbounded because $g(1/k)=k$ for all integers $k\ge 2$. For each $x\in E$, the geometric series gives the sum and its remainder after degree $N$:Since $0<x\le 1/2$,This proves uniform convergence. Although the first term $1/x$ and the sum are unbounded, the late tails have a bound tending to zero independently of $x$. Uniform convergence concerns those tails and does not require the individual functions to be bounded.

For real-valued functions, the Uniform Cauchy criterion applied to partial sums says that $\sum u_n$ has uniform convergence exactly whenSufficiency uses completeness of the real numbers.

Let $M$ be a common bound for the uniformly bounded family of functions $(g_n)$; increase it if necessary so that $M>0$. The assumed uniform absolute convergence of $\sum f_n$ gives $N$ such thatThe triangle inequality now yieldsThe Uniform Cauchy criterion therefore proves uniform convergence of $\sum f_n{g}_n$. The same estimate with $|f_n{g}_n|$ proves uniform absolute convergence as well.