Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 10G (Analysis and Topology)

This is always true. Given $\epsilon >0$, choose $N_A,N_B$ such thaton $A$ for $n\ge N_A$ and on $B$ for $n\ge N_B$. For $n\ge \max (N_A,N_B)$ the same inequality holds at every point of $X=A\cup B$. Hence $f_n\to f$ uniformly on $X$.

This may be false when the limiting functions are unbounded. On $X=\mathbb{R}$, letBoth sequences converge uniformly to their stated limits, butis unbounded for every $n$. Thus the products do not converge uniformly.

For every $(x,y)\in [1,2{]}^2$,Moreover,The bound is independent of $(x,y)$ and tends to zero, so

This is always true. Uniform convergence makes $(f_n)$ a uniformly Cauchy sequence: for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$ and $m,n\ge N$. Letting $x\to x_0$ through points of $A\setminus \{x_0\}$ and using the assumed limits givesThus $(y_n)$ is a Cauchy sequence. Since $Y$ is a complete metric space, $(y_n)$ converges in $Y$.

This is always true. Since $f$ is bounded and $f_n\to f$ uniformly, there are $M$ and $N$ such thatfor every $x\in X$ and $n\ge N$. The continuous function $g$ is uniformly continuous on the compact interval $[-M-1,M+1]$. Therefore, given $\epsilon >0$, a sufficiently small uniform bound on $|f_n-f|$ impliesHence $g\circ f_n\to g\circ f$ uniformly.