I'm having some trouble with the second point of this question and I'm not completely sure that I made the first right so:

For $x\in\mathbb{R} - \{0\}$, let $f_n(x) = \frac{1}{(1 + n^2 + x^2)(\arctan (x^2))^{1/n}}$, $n \in\mathbb{N} - \{0\}$

(i) determine for which $n ∈ \mathbb{N}, f_n$$L^1(\mathbb{R})$

(ii) evaluate $\lim_{n\to \infty}$ $\int_{\mathbb{R}} f_n \,dm$;

For the fist part I notice that the $f_n$ are a.e. continuous so that implies that they are measurable, and that the functions are even, so I can narrow my analysis to $[0, \infty )$.

Then when $x\rightarrow\infty, \, f_n \sim \frac1{x^2\bigl({\pi\over2} \bigr)^{1 \over n}} \le {1 \over {x^2}}$ so I can say that the integral converges for every $n\in\mathbb{N} - \{0\}$

Instead when $x \rightarrow 0$, $f_n \sim \frac1{(1+x^2)x^{2 \over n}} \le {1 \over {x^{2\over n}}}$ that converges for every $n \gt 2 $. Then we came to the second point, here I could not find a an integral majorant to use Lebesgue's dominated convergence theorem and I have not even been able to determine if there is a chain of the type $f_1 \le f_2 \le f_3 \le \cdots \le f_n \le \cdots$ in order to use the monotone convergence theorem in order to pass the limit under sign of integral and evaluate it.

As an addition I think that the integral in the end is zero because $\lim_{n\to \infty} f_n = 0$.

Thank you very much.


Just note that for any fixed $\epsilon \in(0,1)$ then $0\leqslant f_n(x)\leqslant \frac1{(1+x^2)\arctan (\epsilon ^2) }$ for all $x\in \mathbb{R}\setminus (-\epsilon ,\epsilon )$ and all $n\in \mathbb{N}$ so the dominated convergence theorem show us that $$ \lim_{n\to\infty}\int_{\mathbb{R}\setminus (-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda =0 $$ Now note that $|x|/2\leqslant |\arctan x|$ for all $x\in (-\epsilon ,\epsilon )$. Therefore $$ 0\leqslant \int_{(-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda \leqslant \int_{(-\epsilon ,\epsilon )}\frac2{n^2|x|^{2/n}}\mathop{}\!d x =\int_{(0,\sqrt[n]{\epsilon })}\frac4n y^{n-3}\mathop{}\!d y\leqslant \int_{(0,1)}\frac4n\epsilon ^{1-3/n}\mathop{}\!d y $$ Then the dominated convergence theorem show us that $$ \lim_{n\to\infty}\int_{(0,\sqrt[n]{\epsilon })}\frac4n y^{n-3}\mathop{}\!d y=0\implies \lim_{n\to\infty}\int_{(-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda =0 $$ so we conclude that $\lim_{n\to\infty}\int_{\mathbb{R}}f_n\mathop{}\!d \lambda =0$.

  • $\begingroup$ Thank you very much! I understand that a lot of exercise like that are simplified a lot if you are very smart with substitution and increases. Do you have any tip in order to find that or is just exercise and find them at glance? Thank you a lot! $\endgroup$ Oct 28 '20 at 17:29
  • $\begingroup$ @Godel I just tried different things. My first answer was wrong and I deleted it. I starting realizing that something of the form $\frac1{n^2 x^{2/n}}$ cannot have an integrable majorant in a neighborhood of zero, so I tried to write the integral in a different way to see if something work. In short: in this exercise I dont followed a known strategy. $\endgroup$
    – Masacroso
    Oct 28 '20 at 17:37
  • $\begingroup$ Ok got it. Only one last question: when you pass from y to $\epsilon$, is it a necessary observation? when you arrive at the integral in y you can evaluate it easily and after do the limit for n, maybe I'm absolutely wrong but for me it's not the dominated convergence theorem but it's the evaluation of the limit after the integral. As before thank you very much $\endgroup$ Oct 28 '20 at 17:45
  • $\begingroup$ @Godel maybe this clarify your last question $$\int_{(0,1)}\frac4n\epsilon^{1-3/n}\mathop{}\!d y\leqslant \int_{(0,1)}4\mathop{}\!d y=4, \quad \text{ for all }n\geqslant 3$$ I doesn't showed this bound because I think it was trivial. Aside note: notice that its important that $\epsilon\in (0,1)$ to force that $\arctan\epsilon<1$ to make the first bound correct and independent of $n$ $\endgroup$
    – Masacroso
    Oct 28 '20 at 18:10
  • 1
    $\begingroup$ @Godel that is, we have that $\frac4n \epsilon ^{1-3/n}\leqslant 4$ when $n\geqslant 3$ and $\epsilon \in(0,1)$ $\endgroup$
    – Masacroso
    Oct 28 '20 at 18:48

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.