Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 2 / 3F (Analysis II)

Uniform convergence of $f_n$ to $f$ on an interval $I$ means that for every $ϵ>0$ there is $N$ such that $|f_n(x)-f(x)|<ϵ$ for all $n\ge N$ and every $x\in I$. Equivalently ${\sup }_{x\in I}|f_n(x)-f(x)|\to 0$. The same $N$ must work throughout the interval, unlike pointwise convergence.

For completeness, continuity is preserved: choose $N$ with uniform error below $ϵ/3$, and use continuity of $f_N$ at a given $x$ to make $|f_N(y)-f_N(x)|<ϵ/3$. The triangle inequality then gives $|f(y)-f(x)|<ϵ$. Thus each function involved in the integral is continuous on the closed interval and Riemann integrable. The estimateproves convergence of the integrals by uniform convergence. When $a=b$ both integrals are zero.

For each fixed $x\in (0,1)$, $1/(nx)\to 0$, so the only possible limit is zero. But taking $x_n=1/(2n)$ gives $f_n(x_n)=2$ for every $n$. In fact the supremum of $1/(nx)$ over the interval is infinite for each $n$. Thus the convergence is pointwise but not uniform; uniform convergence cannot be inferred from convergence at each fixed point.

The pointwise limit is the constant function $1$. For $0<x<1$, the inequality $1-e^{-u}\le u$ for $u\ge 0$ givesHence the sequence converges uniformly to one on $(0,1)$. Its exact supremum error is $1-e^{-1/n}$, although the supremum is not attained on the open interval. Compactness of the interval is not needed for this uniform convergence bound.