Convergence of $g _ { n } ( x ) = \int _ { 0 } ^ { x }f_n(t)dt $

Question

Let $f _ { n }: \mathbb { R } \rightarrow \mathbb { R }$ be a sequence of $\mathcal C'$ functions, converges to $f$ uniformely, then we have $\int _ { 0 } ^ { x }f_n(t)dt$ converges to $\int_{0} ^ { x } f ( t ) d t ~~\forall x \in \mathbb { R}$, Then what about the convergence of $g _ { n } ( x ) = \int _ { 0 } ^ { x }f_n(t)dt $ to $g ( x ) = \int _ { 0 } ^ { x } f ( t ) d t.$ is it uniform?

Answer 1

It is not. Consider $f(x)= c$ for some constant $c\in\mathbb R$ and $f_n(x)=c+1/n$.

Then $g_n(x)-g(x)=\frac xn$ which doesn't converge to $0$ uniformly on $\mathbb R$.

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EDIT: Additional question: What happens on a closed interval $[0,a]$?

Answer: Then we have for all $x\in[0,a]$, $$|g_n(x)-g(x)|\le\int_0^x |f(x)-f_n(x)|\,\mathrm dx\le x\cdot M_n \le a\cdot M_n$$

where the $M_n$ can be chosen to go to $0$ since the $f_n$ converge uniformly to $f$. So the $g_n$ converge uniformly to $g$ in this case.