4
$\begingroup$

I am currently investigating the convergence of the following function,

$f(x)=\sum\limits_{k=1}^{\infty} \dfrac{x^{k}+\sin(k)}{k^{2}}$

for different "senses". I have shown that $f(x)$ converges uniformly and pointwisely for its domain. All I have left to do is investigate whether it converges in the $L^{p}$ sense. The problem is, I have absolutely no idea what this means! My knowledge on this $L^{p}$ is very minimal and I can not seem to find anything on the www or Rudin. If someone could explain to me what this $L^{p}$ business is it would be greatly appreciated. A few sub-questions I have include:

  • How does it differ from uniform convergence?
  • Any geometrical interpretations?
  • I know that $p\geq1%$. How does every each value of $p$ affect convergence?
  • Are there any relationships between each $L^{p}$?

If anyone knows of some nice references to my specific questions, that also would be greatly appreciated.

Thank you in advanced to all for your time.

$\endgroup$
  • 1
    $\begingroup$ do you know what the $L^p$ norm is $\|f\|_p=(\int|f|^pdx)^{\frac{1}{p}}$ $\endgroup$ – Ellya Apr 9 '14 at 15:25
  • $\begingroup$ Ok, thanks @ellya. That is a start to my understanding. Anymore help? $\endgroup$ – Gustavo Louis G. Montańo Apr 9 '14 at 21:52
  • $\begingroup$ what does f(x) converge to? $\endgroup$ – Ellya Apr 9 '14 at 21:55
  • $\begingroup$ @ellya. It converges, but I do not to what to. The first term, $\frac{x^{k}}{k{2}}$ converges via the ratio test and letting |x| < 1, and the second term converges via the W-M Test. Are you asking for the function that it converges to? I do not know that. Can you help me find it, perhaps? I have proved Uniform and PW convergence via definition of delta-epsilon $\endgroup$ – Gustavo Louis G. Montańo Apr 9 '14 at 22:03
2
$\begingroup$

Well we can see that $|f(x)|=|\sum_{k=1}^\infty\frac{x^k+sin(k)}{k^2}|\leq\sum_{k=1}^\infty|\frac{x^k+sin(k)}{k^2}|\leq \sum_{k=1}^\infty|\frac{x^k}{k^2}|+\sum_{k=1}^\infty\frac{1}{k^2}=\sum_{k=1}^\infty|\frac{x^k}{k^2}|+\frac{\pi^2}{6}$

Now like you said if we restrict to $x\in[0,\frac{1}{2})$ (You'll see why later, then we have $|f(x)|\leq\sum_{k=1}^\infty|\frac{x^k}{k^2}|+\frac{\pi^2}{6}\leq \sum_{k=1}^\infty x^k+\frac{\pi^2}{6}=\frac{1}{1-x}+\frac{\pi^2}{6}$

Now as I said $\|f\|_p=(\int |f|^pdx)^{\frac{1}{p}}$.

Let us look at $L^1$, here $\|f\|_1=\int_0^\frac{1}{2}|f(x)|dx=\int_0^\frac{1}{2}\frac{1}{1-x}+\frac{\pi^2}{6}dx=(-\ln(1-x)+\frac{\pi^2}{6}x)|_0^1=-\ln{\frac{1}{2}}+\frac{\pi^2}{6}=\ln(2)+\frac{\pi^2}{6}\lt\infty$, so $f\in L^1$.

This is all i've thought up so far, but I hope it helps a bit.

edit

I've thought of a much better way to do this.

We as before restrict to $|x|\le 1$, then:

$|f(x)|=|\sum_{k=1}^\infty\frac{x^k+sin(k)}{k^2}|\leq\sum_{k=1}^\infty|\frac{x^k+sin(k)}{k^2}|\leq\sum_{k=1}^\infty\frac{2}{k^2}=\frac{\pi^2}{3}$

Thus $\int_{-1}^{1}|f(x)|^pdx\leq\int_{-1}^{1}(\frac{\pi^2}{3})^pdx=2(\frac{\pi^2}{3})^p\lt\infty$, Thus $f\in L^p([-1,1])$

$\endgroup$
  • $\begingroup$ Hey @ellya, yes that has made more sense. A few questions. What you did in the first 4 lines, what did you achieve? Did you show that my original function converges to $\sum_{k=1}^\infty|\frac{x^k}{k^2}|+\frac{\pi^2}{6}$? If so, I can not see why. Sure, I know the function converges, but would not be satisfied to conclude that it converges to that. Now, the domain is [-1,1]. And I am curious to know why you only went from 0 -> (1/2).How about L2 and L3 and so on? Thanks ! $\endgroup$ – Gustavo Louis G. Montańo Apr 9 '14 at 23:04
  • $\begingroup$ Firstly, what I found was an upper bound for f, which allowed me to show that f converges under the $L^1$ norm. I took $(0,1/2)$ literally because it simplifies this integral, Im sure with more effort you can extend the interval. As for L^p in general, the integral just becomes a bit more complicated, as we take the pth power in the integral. $\endgroup$ – Ellya Apr 9 '14 at 23:11
  • $\begingroup$ I see. With the $L^{p}$ is it possible though? Because I am required to see if it converges in the $L^{p}$ sense. So, how can I come to a final conclusion? Any hint - I don't expect you to do it for me. $\endgroup$ – Gustavo Louis G. Montańo Apr 9 '14 at 23:18
  • $\begingroup$ Try what I did, firstly on (0,1/2) BUT do the integral of $|f|^p$ and see what happens,hopefully it will be finite, after that try making the interval larger. $\endgroup$ – Ellya Apr 9 '14 at 23:21
  • $\begingroup$ Also what was the domain of f meant to be? Its not in your post. $\endgroup$ – Ellya Apr 9 '14 at 23:24

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.