Uniform convergence integration question EDIT: Possibly major detail I forgot to mention: $f$ is $C^1$.
The question: Show that there exist constants $c_n>0$ (independent of $f$) such that
the sequence of functions
$$g_n(x)=c_n \int_{-\pi/n}^{\pi/n} f(x+t)\sin(nt)\,dt$$
converges to $f'(x)$, uniformly on every bounded interval $[-M,M]\subset \mathbb{R}$.
I find this question extremely weird. No other uniform convergence problem I remember seeing in this class has an integral converging to a derivative. Also, the integral of two unrelated functions would suggest convolution, but neither has a compact support, as both f(x+t) and $\sin(nt)$ are supported over all of $\mathbb{R}$. So, I'm really just not sure where to proceed. Any help would be great. Martin.
 A: The trick is an old friend: integration by parts.
$$\int_{-\pi/n}^{\pi/n} f(x+t)\sin(nt) \, dt = \frac 1 n (f(x + \pi/n) - f(x - \pi/n)) + \frac 1 n \int_{-\pi/n}^{\pi/n} f'(x+t)\cos(nt) \, dt.$$
The first term is suspiciously similar to a difference quotient, so from here one can ponder what $c_n$ should be. I can clarify any additional details, but IBP is the crux of the argument.
EDIT to include the rest of the proof: now we have
$$\frac{n}{2\pi}(f(x + \pi/n) - f(x - \pi/n)) + \frac{n}{2\pi} \int_{-\pi/n}^{\pi/n} f'(x+t)\cos(nt) \, dt$$
and we need to show that the first term converges uniformly to $f'(x)$ and the second term converges uniformly to $0$.
Let $\epsilon > 0$, and put $D(x, n) = (n/(2\pi))(f(x + \pi/n) - f(x - \pi/n))$. Since $f'$ is uniformly continuous on a bounded interval, there is $N \in \mathbb N$ such that for any $t_1, t_2 \in [-M, M]$ such that $|t_1 - t_2| < 2\pi/N$ implies that $|f'(t_1) - f'(t_2)| < \epsilon.$ Let $n \geq N$. By the Mean Value theorem, there exists $t \in [x - \pi/n, x + \pi/n]$ such that $D(x,n) = f'(t)$. Therefore,
$$ |D(x, n) - f'(x)| \leq |D(x,n) - f'(t)| + |f'(t) - f'(x)| = |f'(t) - f'(x)| < \epsilon$$
and the difference quotient uniformly converges to $f'$ on $[-M,M]$.
Finally, we show that $\frac{n}{2\pi} \int_{-\pi/n}^{\pi/n} f'(x+t)\cos(nt) \, dt \to 0$ uniformly. By the change of variables $u = nt$, we get that
$$ \frac{n}{2\pi} \int_{-\pi/n}^{\pi/n} f'(x+t)\cos(nt) \, dt = 2\pi\int_{-\pi}^{\pi} f'(x + u/n)\cos(u) \, du.$$ Pointwise convergence to $0$ is now pretty clear since $f'(x + u/n)\cos(u) \to f'(x)\cos(u)$ uniformly, so for each $x \in [-M,M]$, $$\lim_{n\to \infty} \int_{-\pi}^{\pi} f'(x + u/n)\cos(u) \, du = \int_{-\pi}^{\pi} \lim_{n\to\infty} f'(x + u/n) \cos(u) \, du = f'(x) \int_{-\pi}^\pi \cos(u) \, du = 0.$$
It turns out that if $I,J$ are compact intervals and $g_n : I \times J \to \mathbb R$ is a sequence of continuous functions converging to $g$ uniformly, then $f_n(x) = \int_J g_n(x,t)\, dt$ converges to $f(x) = \int_J g(x,t) \, dt$ uniformly. Namely, if $g_n(x,u) = f'(x + u/n) \cos(u)$ we have that $\int_{-\pi}^\pi g_n(x,u) \, du \to 0$ uniformly.
