I'm trying to evaluate these integrals using Convergence Theorems, but I'm not really sure how to go about it. Here are the integrals:

  • For $\phi$ bounded and continuous, $\psi\in L^1(m)$, $\lim_{n\to\infty}\int_{\mathbb R}\phi(x/n)\psi(x)dm(x)$
  • For $\phi$ continuous and compactly supported, $\psi\in L^1(m)$, $\lim_{n\to\infty}\int_{-\infty}^\infty \phi(nx)\psi(x)dm(x)$
  • $\lim_{n\to\infty}\int_{-\infty}^\infty \frac{n}{x}\sin(x/n)e^{-\lvert x\rvert}dm(x)$
  • $\lim_{n\to\infty}\int_{[0,1]}(1+nx^2)(1+x^2)^{-n}dm(x)$

$m$ denotes the Lebesgue measure, and $L^1(m)$ denotes the set of integrable functions with respect to the Lebesgue measure.

I think for the first two, I need to find a dominating function for the integrand and then find the limit of the integrand, and for the last two, I believe I need to show that they are monotone increasing and find the limit of the integrand, but I haven't been able to come up with dominating functions/proof that the sequence of functions are monotone increasing.

I would really love to get some help.


Here are some hints:

  1. Use the fact that $\phi$ is bounded, say $|\phi|\le M$ and $M\psi$ as a dominating function. Apply dominated convergence.

  2. Similar idea to 1. Since $\phi$ is compactly supported, if $x \ne 0$, what is the value of $\phi(nx)$ for large $n$?

  3. Apply 1. to $\phi(x) = \sin x/x$ and use the fact that $\sin x/x\to 1$ as $x\to 0$.

  4. Notice that the denominator $(1+x^2)^n \ge 1 + nx^2 + n(n-1)x^4/2$ by the binomial theorem holds for each $n$. Find an appropriate dominating function.
  • $\begingroup$ Couple questions: 1. continuous and compactly supported should imply boundedness? 2. Could you please elaborate a bit more regarding the 3rd question? $\endgroup$ – Mog Oct 25 '18 at 20:41
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    $\begingroup$ @pilotmath: If a function is continuous and supported on a compact set, then it is bounded (since it attains its maximum and minimum in its support). For the third question, we are recognizing that $\phi(x) = \frac{\sin(x)}{x}$ is a bounded and continuous function (in fact, $|\phi(x)| \le 1$ for all $x\in\Bbb R$). My comment about the limit as $x\to0$ of $\phi(x)$ was just so that you see that $\phi$ is bounded, since the only potential problem point is at $x = 0$, where the denominator goes to $0$. Since $e^{-|x|}$ is in $L^1$, your first problem applies directly to this special case. $\endgroup$ – Alex Ortiz Oct 25 '18 at 21:18
  • $\begingroup$ @AOrtize Thank you for your answer. Just to clarify, so having $x/n$ has no effect on the problem, rather I can just bound it by 1, and simply compute the integral of $e^{-\lvert x\rvert}$? Also the answer for 1 would then be, $M\int_{\mathbb R}\phi(0)\psi(x)dm(x)$, and the answer for 2 would be $0$, as Lebesgue measure of a singleton is $0$, right? $\endgroup$ – Mog Oct 25 '18 at 21:30
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    $\begingroup$ @pilotmath: The $x/n$ matters, and it can't be bounded by $1$ since $x$ can take any value in $\Bbb R$. Your answer for 1 is not quite right, but it's very close. You have an extra factor of $M$. Your answer for 2 is correct, but for the wrong reason. $\endgroup$ – Alex Ortiz Oct 25 '18 at 21:35
  • $\begingroup$ @AOrtize right, I understand 1 now, made an algebra error. For 2, I have $\phi(0)\int_{\{0\}}\psi(x)dm(x)$ which should equal to 0, I believe. Please correct me if I'm wrong. I'm still a bit confused about 3, what should $n/x\sin(x/n)$ be bounded by? And once I find the bound, the integral itself should be 0 as $\lim n/x\sin(x/n)$ equal to 0, right? I apologize for asking a lot of questions, I just want to make sure I understand every step. $\endgroup$ – Mog Oct 25 '18 at 21:44

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