Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 3E (Analysis II)

A sequence of functions $f_n:A\to \mathbb{R}$ converges uniformly to $f$ when, for every $\epsilon >0$, there is an $N$ such thatfor every $x\in A$. Equivalently, $\Vert f_n-f{\Vert }_{\infty }\to 0$ in the supremum norm.

Yes. Fix $x_0\in A$ and $\epsilon >0$. Uniform convergence gives $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in A$. Since the continuous function $f_N$ is continuous at $x_0$, there is a $\delta >0$ such that $|x-x_0|<\delta$ implies $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality then givesThis proves the uniform limit theorem: $f$ is continuous.

No. For each $x>0$, $e^{-nx}\to 0$, whereas $e^{-n0}=1$ for every $n$. The pointwise limit on $[0,1]$ is thereforeThis limit is not continuous at $0$. Since every $f_n(x)=e^{-nx}$ is continuous, the uniform limit theorem rules out uniform convergence. Directly, $|f_n(x)-f(x)|=e^{-nx}$ for $x>0$, whose supremum over $(0,1]$ is $1$ for every $n$.

No. DefineEach $f_n$ is a differentiable function on $[-1,1]$, whileThus $f_n$ converges uniformly to the absolute value function $|x|$, which is not differentiable at $0$. Uniform convergence alone therefore does not preserve differentiability.