Let [tex]f(x)=\sum_{n=1}^{\infty}\frac{(nx)}{n^2}[/tex], where (x) denotes the fractional part of x. [tex]|\frac{(nx)}{n^2}| \leq \frac{1}{n^2}[/tex], so [tex]\sum \frac{(nx)}{n^2}[/tex] converges uniformly to f on all of R. (x) is discontinuous precisely at the integers, so [tex]s_k(x) = (x) + \frac{(2x)}{2^2} + ... + \frac{(kx)}{k^2}[/tex] is discontinuous precisely when [tex]x = \frac{p}{m}[/tex], where m = 1,...,k, and p is any integer. It follows that if [tex]x_0[/tex] is irrational, then [tex]\lim_{x \rightarrow x_0}s_n(x) = s_n(x_0)[/tex] for all n. So by theorem 7_11 in Rudin, [tex]\lim_{x \rightarrow x_0}f(x) = \lim_{n \rightarrow \infty}s_n(x_0) = f(x_0)[/tex], which shows f is continuous at every irrational point.(adsbygoogle = window.adsbygoogle || []).push({});

Another thing to add is that it is clear f(x+1) = f(x) for all x. It is also clear by uniform convergence that f is Riemann integrable on any bounded interval, since each s_n is as such. That's pretty much most of Problem 10 in Chapter 7, but not all.

Is the limit function discontinuous on Q (the rationals)? Or equivalently, by the f(x+1)=f(x) property, is f discontinuous at all the rationals in [0,1]? I don't see off hand the direct reason WHY the limit function is discontinuous, despite the fact that every rational number q is a discontinuity point of s_n, for all n >= some N. (In particular, the problem asks to show the discontinuous points are a countable dense subset, which would be implied if that subset is Q...) I.e., just because g_n(x) = 1/n, x rational, g(x) = 0, x irrational, is discontinuous everywhere, doesn't imply it's uniform limit function g(x)=0 is discontinuous anywhere.

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# A function discontinuous only on Q

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