Let $f_n$ on $[-\pi,\pi]$ be: $$f_n(x)=\frac{\sin nx}{n}$$ Let $f$ be the pointwise limit of $f_n$
Denote $F(x)=\int_{-\pi}^x f(y)dy$ and $F_n(x)=\int_{-\pi}^x f_n(y)dy$.
On $[-\pi,\pi]$:
a)Does $F_n \rightarrow F$ uniformly?
b)Does $f'_n \rightarrow f'$ uniformly?
My Solution:
First I found the pointwise limit of $f_n(x)$ to be $f=0$. I then said that $f_n \rightarrow f$ uniformply since $$\lim_{n\rightarrow \infty} \sup | \frac{ \sin (nx)}{n}-0|=0<\epsilon$$
Hence, I thought about interchanging the limit and integral signes, to show that both a and b are true. However, this is not necessarily true, e.g. my answer to b is false, why can't I interchange?
a) $F_n(x)=\int_{-\pi}^x f_n(y)dy=\int_{-\pi}^x \frac{ \sin (nx)}{n}dy=-\cos (\pi x)/n^2+\cos (-\pi x)/n^2=0$ which agrees with the zero integral.
b)$\frac{ \sin (nx)}{n}'=\cos (nx)$ but $\sup |\cos (nx)-0|\ne 0$ hence it's false.
$$ |F_n(x)-0|=|\int_{-\pi}^x f_n(y)dy \ | \leq \int_{-\pi}^x |f_n(y)|\leq \int_{-\pi}^x \frac{1}{n}dy \leq 2\pi/n$$ We wee that independent of $x$, the limit goes to zero. Also note that $F=0$ as $f=0$ for all $x$.
Here you could interchange because convergence $F_n \rightarrow F$ is uniform. Derivatives do not converge even pointwise, take $x=\pi/2$ for instance. So, we don't expect anything, any relation between $f_n'$ and $f'$. Derivatives have zero respect for convergence (even uniform convergence), and it is plausible, take a vary rigid saw-shaped function, positive and bounded. Then multiply be $\frac {1}{n}$'s to have a uniformly converging seq about whose derivative you can say nothing!