Let $F(x) = (1-x^2) (H(x+1) -H(x-1))$. Let $g$ be a continuous function on the interval $[-1, 1]$. Find $$ \text{lim} _{n \rightarrow \infty}\ \frac{3}{4} \int_{-1}^1 nF(nx)g(x) \ \text{dx}$$
$H(x)$ is Heaviside step function.
I think that limit would be $g(0)$. But I can not prove it. Please help me.
Sketch: Observe that \begin{align} H(x+1)-H(x-1) = \chi_{[-1, 1]}(x) \end{align} which means \begin{align} F(nx) = (1-n^2 x^2)\chi_{[-1, 1]}(n x) = (1-n^2x^2)\chi_{[-1/n, 1/n]}(x). \end{align} then it follows \begin{align} \int^1_{-1}nF(nx)g(x) dx = \int^{1/n}_{-1/n}n(1-n^2 x^2)g(x)\ dx = \int^1_{-1} (1-z^2) g(\frac{z}{n})\ d z. \end{align} Hence it follows that \begin{align} \lim_{n\rightarrow \infty} \frac{3}{4}\int^1_{-1}(1-z^2)g(z/n) dz = \frac{3}{4}g(0) \int^1_{-1}(1-z^2)dz = g(0). \end{align}
I will let you fill out the rigorous detail about the interchanging of limit and integration.