Given two functions, $f, g \in C_c^{\infty}[-1,1]$ is the set of smooth functions that are compactly supported on $[-1,1]$, I want to evaluate: $$ \lim_{n \to \infty} \int_{-n}^n f(x/n)g(x) \text{d}x. $$
If we let $h_n(x) = f(x/n)g(x)$, then $h_n(x) \to f(0) g(x)$ pointwise, can I find a dominating function and use the dominated convergence theorem here? I'm a bit confused since the limits of integration are variable, but since the support of both functions is on the interval $[-1,1]$ can I just ignore this?
Under the hypothesis that $\text{supp}(g)\subseteq[-1,1]$, then for all $n \geq 1$ $$ \int_{-n}^n f(x/n)g(x) dx = \int_{-1}^{1} f(x/n)g(x) dx, $$ and the answer is somewhat trivial. For a matter of commodity, suppose that $f(x) \geq 0$ and $g(x)\geq 0$ for all $x \in \mathbb{R}$. Since $f$ is continuous on a compact set, we can define $m = \min_{x\in[-1,1]}\{f(x)\}$ and $M = \max_{x\in[-1,1]}\{f(x)\}$. Then for all $n>1$ $$ m\int_{-1}^{1} g(x) dx\leq \int_{-1}^{1} f(x/n)g(x) dx \leq M\int_{-1}^{1} g(x) dx. $$ A similar answer applies if $\text{supp}(g)\subseteq[-k,k]$ for some $k \in \mathbb{N}$.
Some trouble arises if $\text{supp}(g)$ is not compact. For instance, let $g(x)=1$ for all $x \in \mathbb{R}$ and let $f(x)$ be your favorite smooth function such that
Then $$ \int_{-n}^n f(x/n)g(x) dx \geq \int_{-n/2}^{n/2} f(x/n)g(x) dx = \int_{-n/2}^{n/2} 1 \cdot 1 dx = n. $$ As a consequence, the sequence $n \mapsto \int_{-n}^{n} f(x/n)g(x) dx$ diverges.