1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.