Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 1 / 1E (Analysis II)

No. Consider the triangular functions $f_n(x)=\max (1-n|x|,0)$. The absolute-value and positive-part maps are Lipschitz continuous, so $|f_n(x)-f_n(y)|\le n|x-y|$. Each $f_n$ is therefore uniformly continuous on the whole real line. For $x=0$ its value is always one; for every fixed $x\ne 0$ it eventually vanishes. Hence its pointwise limit isThis limit is not continuous at zero: along $x_j=1/j\to 0$, $f(x_j)=0\nrightarrow 1=f(0)$. Individual uniform continuity does not provide a modulus uniform in $n$.

Yes; in fact the limit is uniformly continuous. Given $\epsilon >0$, uniform convergence provides one $N$ with ${\sup }_x|f_N(x)-f(x)|<\epsilon /3$. Uniform continuity of this particular $f_N$ provides $\delta >0$ such that $|x-y|<\delta$ implies $|f_N(x)-f_N(y)|<\epsilon /3$. The triangle inequality then givesThis same $\delta$ works at every point of $\mathbb{R}$, proving uniform continuity and therefore continuity. The use of one global $N$ is exactly what fails for mere pointwise convergence.