Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 12F (Analysis II)

The sequence $(f_n)$ converges uniformly to $f$ on $A$ when, for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$.

The function $f$ is uniformly continuous when, for every $\epsilon >0$, there is $\delta >0$ such that $x,y\in A$ and $\Vert x-y\Vert <\delta$ imply $|f(x)-f(y)|<\epsilon$. The same $\delta$ works throughout the domain.

True. Given $\epsilon >0$, uniform continuity supplies $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. For all sufficiently large $n$, $1/n<\delta$, and thenThis is precisely uniform convergence of $f_n$ to $f$.

False. Take the uniformly continuous function $f(x)=x$. Thenwhich is unbounded as a function of $x$ for every fixed $n$. Thus $g_n$ does not converge uniformly to $f^2$ on $\mathbb{R}$.

The closed bounded set $A\subset {\mathbb{R}}^m$ is compact by the Heine-Borel theorem. Choose a convergent subsequence $x_{n_j}\to x_0$. For fixed $N$ and sufficiently large $j$, monotonicity givesContinuity and passage to the limit give $g_N(x_0)\ge \delta$ for every $N$, hence ${\lim }_N{g}_N(x_0)\ge \delta$.

For the deduction, set $g_n=f_n-f$. These functions are continuous, decrease pointwise, and tend to zero. If convergence were not uniform, some $\delta >0$ would admit $x_n$ with $g_n(x_n)\ge \delta$. The result just proved would produce $x_0$ with $\lim g_n(x_0)\ge \delta$, contradicting pointwise convergence. Therefore $f_n\to f$ uniformly. This is Dini's theorem.