Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 4G (Analysis II)

Uniform convergence means that for every $\epsilon >0$ there is an integer $N$, independent of $x$, such that $|F_n(x)-F(x)|<\epsilon$ for all $n\ge N$ and every $x\in (0,1)$. Equivalently ${\sup }_{0<x<1}|F_n(x)-F(x)|\to 0$. In pointwise convergence, the required $N$ may depend on the chosen $x$.

At every fixed $x$, $F_n(x)=e^x/n\to 0$. Moreover ${\sup }_{0<x<1}{e}^x/n=e/n\to 0$. Thus the pointwise limit is zero and convergence is uniform.

For every fixed $x>0$, $n{x}^2\to \infty$, so the pointwise limit is zero. But ${\sup }_{0<x<1}{e}^{-n{x}^2}=1$ for every $n$, since $x$ can approach zero. Hence convergence is not uniform. For an explicit moving-point test take $x_n=n^{-1/2}$ for $n>1$, which gives $F_n(x_n)=e^{-1}$.

The finite geometric series gives $F_n(x)=(1-x^{n+1})/(1-x)$, so the pointwise limit is $F(x)=1/(1-x)$. The error is $x^{n+1}/(1-x)$, unbounded as $x\uparrow 1$ for every fixed $n$. Thus convergence is not uniform on $(0,1)$. It is uniform on any smaller interval bounded away from one, since the error is at most $b^{n+1}/(1-b)$ for $0<x\le b<1$.