If $g \ge 0$ vanishing outside $(-1,1)$ with integral $1$ what is $\lim_{t\to 0}\frac 1t\int_{-\infty}^{\infty} g\left(\frac xt\right)f(x)\,dx$?

Question

Let $g:\mathbb R \to \mathbb R$ be a continuous function such that

1. $g(x)\ge 0, \forall x \in \mathbb R$,
2. $g(x)=0, \forall |x| \ge 1$ and
3. $\int_{-\infty}^{\infty} g(x)\, dx=1 $

and let $f: \mathbb R \to \mathbb R$ be any continuous function. Then how to evaluate $$\lim_{t\to0}\dfrac 1t\int_{-\infty}^{\infty} g\left(\dfrac xt\right)f(x)\,dx$$ From given, it is indeed true that $\int_{-1}^1g(x)\,dx=1$ and the given limit is equal to (assuming it exists) $$\lim_{t \to 0+}\int_{-t}^{t} g\left(\dfrac xt\right)f(x)\,d\left(\dfrac xt\right)=\lim_{t \to 0+}\int_{-1}^1g(z)f(tz)\,dz$$ but I cannot proceed further, please help. Thanks in advance.

Answer 1

By the mean value theorem for definite integrals

If $f : [a, b] \to \mathbb R$ is continuous and $g$ is an integrable function that does not change sign on $[a, b]$, then there exists $c$ in $(a, b)$ such that $$\int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx$$

Your functions satisfy these assumptions, so for $(a,b)=[-1,1]$ you get that $$\int_{-1}^{1}g(z)f(tz)dz=f(tc)\int_{-1}^{1}g(z)dz=f(tz)$$ Taking the limit $t\to 0$, the RHS converges due to continuity of $f$ to $f(0)$.