$\lim_{n \to \infty} n \int_{0}^{100}f(x)g(nx)dx=f(0)$

Question

Let $g: \mathbb{R} \to \mathbb{R}$ be a continuous function with $g(y)=0$ for all $y \notin [0,1]$ and $\int_{0}^{1}g(y)dy=1$. Let $f: \mathbb{R} \to \mathbb{R}$ be a twice differentiable function. Then $$M:=\lim_{n \to \infty} n \int_{0}^{100}f(x)g(nx)dx=f(0)$$

Sol: I let $nx=t$, and split $M$ into two parts: $$M=\lim_{n \to \infty} \left ( \int_{0}^{1}f\left ( \frac{t}{n} \right )g(t)dt+\int_{1}^{100n}f\left ( \frac{t}{n}\right) g(t)dt \right )$$ If I can move the limit inside the integral, the first part becomes $f(0)$ and the second part is $0$ since $g$ is $0$ outside $[0,1]$. This discusses the conditions under which it can be done. Is this correct?

Answer 1

It suffice to prove $$\lim_{n \to \infty} \int_0^1 f(\frac tn)g(t) \,dt=f(0)$$ Notice that $$f(0)=\int_0^1 f(0)g(t) \,dt$$ What we need to prove is $$\lim_{n \to \infty} \int_0^1 (f(\frac tn)-f(0))g(t) \,dt=0$$ This can be done since $f$ is uniform continuous and $g$ is bounded .