suppose we have a sequence of continuous functions $(f_n)_{n\geq 1}$ with $f_n:[0,\infty)\rightarrow \mathbb{R}$ such that $\int_{0}^{\infty}\vert f_n(x) \vert<\infty$ for all $n\in\mathbb{R}$. Further, let $\lim\limits_{n\rightarrow \infty}f_n(x) = 0$ for all $x\geq 0$.

Is it true that $$\lim\limits_{n\rightarrow \infty}\int_{0}^{\infty}f_n(x)\cdot g(x) dx = 0,$$

if $g:[0,\infty)\rightarrow \mathbb{R}$ is a continuous and absolutely integrable function (i.e. $\int_{0}^{\infty}\vert g(x) \vert<\infty$)? Can I interchange the order of integration and the limit? (If needed, we may also assume that $\int_{0}^{\infty}\vert f_n(x)\cdot g(x) \vert<\infty$ for all $n$).

Best regards


If $f_n\to 0$ uniformly then the answer is affirmative (see the comments below). In general the answer is no, however. Here is a counter example. Consider the sequence $$f_n(x)=e^{-(x-n)^2+x}\;,$$ and let $g(x)=e^{-x}$. Then $f_n\to 0$ pointwise but not uniformly on $[0,\infty)$. By translation invariance of $dx$, we have $$\lim_{n\to \infty}\int_0^{\infty}e^{-(x-n)^2+x}e^{-x}dx=\lim_{n\to \infty}\int_0^{\infty}e^{-(x-n)^2}dx =\lim_{n\to\infty}\int_{-n}^{\infty}e^{-y^2}dy=\sqrt{\pi}$$

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    $\begingroup$ You do not really need uniform convergence. It suffices to have $|f_n (x)| \leq C $ for all $n $ and (almost) all $x $. $\endgroup$ – PhoemueX Mar 17 '17 at 8:26
  • $\begingroup$ Yes, I did not want to say anything against your counterexample. I only objected to the sentence "unless $f_n \to 0$ uniformly, the answer is no". This sounds as if the claim was always false, unless we have uniform convergence. But in fact, it suffices to have $|f_n(x)| \leq C$ (which is implied by uniform convergence). (BTW: +1) $\endgroup$ – PhoemueX Mar 17 '17 at 9:53
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    $\begingroup$ But if $|f_n (x)| \leq C$ for all $x, n$, then $|f_n (x) \cdot g(x)| \leq C \cdot |g(x)|$, where the right-hand side is in $L^1$. Also, $f_n (x) \cdot g(x) \to 0$ pointwise, so that the dominated convergence theorem yields $\int f_n (x) \cdot g(x) \, dx \to 0$. $\endgroup$ – PhoemueX Mar 17 '17 at 12:12
  • $\begingroup$ @PhoemueX Yes you're right. I made a mistake. My $f_n$'s are not uniformly bounded...thanks for pointing that out. $\endgroup$ – user113529 Mar 17 '17 at 12:58

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