Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 2G (Analysis II)

The sequence has uniform convergence on $X$ if there is a function $f:X\to \mathbb{R}$ such thatThe same $N$ must work for every $x$. Equivalently, the supremum of the error over $X$ tends to zero. In pointwise convergence, $N$ is allowed to depend on $x$.

For each fixed $0<x<1$, $x^n\to 0$, so the pointwise limit is $f(x)=1/(1-x)$. Howeverfor every fixed $n$. Hence the error has infinite supremum on $(0,1)$ and the convergence is not uniform. This does not contradict uniform convergence on any smaller closed interval $[0,r]$, $r<1$, where the error is at most $r^n/(1-r)$.

For $0<x<1$, $|x^k/k^2|\le 1/k^2$, and the numerical series ${\sum }_{k\ge 1}{k}^{-2}$ converges. The Weierstrass M-test therefore gives uniform convergence to $f(x)={\sum }_{k\ge 1}{x}^k/k^2$. More explicitly,The bound is independent of $x$, including points arbitrarily close to one.

For each fixed real $x$, $x/n\to 0$, so the pointwise limit is zero. But ${\sup }_{x\in \mathbb{R}}|x/n|=\infty$ for every $n$. Alternatively, $x=n$ gives error one whatever $n$ is. Thus the convergence is not uniform on $\mathbb{R}$. It is uniform on each bounded interval, where $|x/n|\le M/n$.

The pointwise limit is zero, including at $x=0$. Differentiation givesso the maximum on $[0,\infty )$ is attained at $x=1/n$ and equals $1/(en)$. Thereforeand the convergence is uniform, even though the domain is unbounded.