Dominated convergence with variable limit

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?

• Looks like a typo in the question somewhere, $g$ needs to be defined outside the interval $[-1,1]$ for the integral to make sense. Of course, I don't really know what $C_c^{\infty}[-1,1]$ means (what is that $c$ subscript?) – Michael Oct 6 '18 at 7:34
• @Michael I've clarified in the question – dimebucker Oct 6 '18 at 7:38
• I don't think anyone would write a problem like that and assume we know that $g$ is supposed to be zero outside. – Michael Oct 6 '18 at 7:39
• @Michael Sorry I thought the $c$ subscript notation was standard – dimebucker Oct 6 '18 at 7:40
• Standard for what? – Michael Oct 6 '18 at 7:40

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

• $$f(x) \geq 0$$ for all $$x \in\mathbb{R}$$;
• $$f(x)=1$$ for all $$x\in\mathbb{R}$$ with $$|x|\leq1/2$$ and
• $$f(x)=0$$ for all $$x\in\mathbb{R}$$ with $$|x|\geq0$$.

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.

• Your second line of inequalities need fixing if $g$ is allowed to be negative. Though, I think the spirit of this answer is similar to my above comment that $f$ is bounded. – Michael Oct 6 '18 at 23:03
• Thanks, I added the hypotheses that $f, g \geq 0$. – Emanuele Bottazzi Oct 7 '18 at 16:59
• Looks good (+1). – Michael Oct 8 '18 at 2:29