I was given this exercise as a homework assignment, and I couldn't solve it, any help would be appreciated-

Define $$ f_n(x) = \sum_{k=0}^n\frac{1}{2^k(x-x_k)} $$ for some sequence $x_n$. Show that $f_n(x)$ is a.e convergent in $\mathbb{R}$, as $n\to\infty$.

Thanks in advance!

  • 1
    $\begingroup$ You likely mean $$ f_n(x) = \sum_{k=0}^n\frac{1}{2^k(x-x_k)}? $$ $\endgroup$ – gt6989b Dec 31 '17 at 20:40
  • $\begingroup$ Oh yes, thank you. I'll fix that $\endgroup$ – user223740 Dec 31 '17 at 20:40
  • $\begingroup$ Bad x's, $x=x_k$ or x is a limit of a subsequence of ${x_k}$. These sets are countable. You need to show that using $\frac{1}{2^k} $, at all other points the series converges. $\endgroup$ – herb steinberg Dec 31 '17 at 22:28
  • $\begingroup$ @herbsteinberg What if $\{ x_n \}$ enumerates the rational numbers? Wouldn't every irrational number have a subsequence converging to it? Are the sequences perhaps defined to be convergent? $\endgroup$ – Trevor Norton Dec 31 '17 at 23:11
  • $\begingroup$ @Trevor Norton - your comment is valid. Proving the original question seems to be harder than I thought. $\endgroup$ – herb steinberg Jan 1 '18 at 17:13

Hint: One of those standard little tricks shows that given $a$, you have $$\sum_k\frac1{2^{k/2}|x-x_k|^{1/2}}<\infty$$for almost every $x\in [a,a+1]$.

Solution: (If you want to figure it out yourself then don't hover the mouse over the yellow box...)

You can verify that $\int_a^{a+1}|x-x_k|^{-1/2}\,dx\le c$ by calculus. So $$\int_a^{a+1}\sum 2^{-k/2}|x-x_k|^{-1/2}\,dx=\sum 2^{-k/2}\int_a^{a+1}|x-x_k|^{-1/2}\,dx<\infty,$$ hence $\sum 2^{-k/2}|x-x_k|^{-1/2}<\infty$ almost everywhere on $[a,a+1]$, hence almost everywhere on $\mathbb R$. But $\sum|\alpha_k|<\infty$ implies $\sum|\alpha_k|^2<\infty$.

  • $\begingroup$ Thank you!! Very helpfull $\endgroup$ – user223740 Jan 1 '18 at 7:47
  • $\begingroup$ So you got it, great. It's a curious thing. Say $a_j>0$. Then convergence of $\sum a_j$ is a much stronger condition than convergence of $\sum a_j^2$: curious that here we show $\sum a_j^2$ converges by showing that $\sum a_j$ does. $\endgroup$ – David C. Ullrich Jan 1 '18 at 13:49
  • $\begingroup$ Yes, that's really surprising. I wonder if there is a straightforward proof for this casa $\endgroup$ – user223740 Jan 1 '18 at 17:17
  • $\begingroup$ @NoamPirani Perhaps more straightforward, less of a "trick". also less fun: Let $E_k=\{x:2^{-k}|x-x_k|^{-1}>2^{-k/2}\}$. Then $\sum m(E_k)<\infty$, hence almost every $x$ lies in only finitely many $E_k$. $\endgroup$ – David C. Ullrich Jan 1 '18 at 17:45
  • $\begingroup$ Nice! You deduce that a.e x is only in finitley many $E_k$ by borel-cantelly, right? $\endgroup$ – user223740 Jan 2 '18 at 19:57

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