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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?

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    $\begingroup$ 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?) $\endgroup$ – Michael Oct 6 '18 at 7:34
  • $\begingroup$ @Michael I've clarified in the question $\endgroup$ – dimebucker Oct 6 '18 at 7:38
  • $\begingroup$ I don't think anyone would write a problem like that and assume we know that $g$ is supposed to be zero outside. $\endgroup$ – Michael Oct 6 '18 at 7:39
  • $\begingroup$ @Michael Sorry I thought the $c$ subscript notation was standard $\endgroup$ – dimebucker Oct 6 '18 at 7:40
  • $\begingroup$ Standard for what? $\endgroup$ – Michael Oct 6 '18 at 7:40
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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.

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  • $\begingroup$ 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. $\endgroup$ – Michael Oct 6 '18 at 23:03
  • $\begingroup$ Thanks, I added the hypotheses that $f, g \geq 0$. $\endgroup$ – Emanuele Bottazzi Oct 7 '18 at 16:59
  • $\begingroup$ Looks good (+1). $\endgroup$ – Michael Oct 8 '18 at 2:29

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