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Suppose I have a series $\sum_{n = 0}^{\infty} f_{n}(x)$ which converges absolutely to a function $f(x)$. Does the series converge uniformly to $f(x)$? I want to say this follows from Dini's Theorem, but I can't seem to see how.

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Try $f_n(x):=x^n(1-x)$ on $\left[0,1\right]$ (if I understand the question correctly). –  Davide Giraudo Oct 26 '11 at 17:09
    
You can also try $f_n = \frac{x}{n^2}$ on $R$. –  N. S. Oct 26 '11 at 17:31

3 Answers 3

up vote 5 down vote accepted

There are easy counterexamples to the statement, which others have already supplied.

But let me say what mistake I think you're making. Dini's Theorem has a lot of fine print: in particular, it requires the pointwise limit function to be continuous. You can see that this hypothesis is not met in the above counterexamples. Conversely, if each $f_n$ and also $\sum_{n=0}^{\infty} |f_n(x)|$ were continuous, then indeed you could apply Dini's Theorem to see that the convergence of $\sum_n |f_n(x)|$ is uniform, from which it follows that the convergence of $\sum_n f_n(x)$ is uniform, say via the Cauchy criterion.

You have my sympathies for forgetting this hypothesis of Dini's Test. I still have a set of lecture notes up on my webpage where I state Dini's Test incorrectly (I can't find the tex file to change it!), and when I mentioned it in a second course more recently, I made my way through the statement and the proof (!) without making explicit the need for continuity of the limit function. Finally I got it right: see for instance $\S III.1.4$ of these notes.

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+1, if only for the phrase a lot of fine print. –  Did Oct 29 '11 at 10:32

The example given by Davide Giraudo works well. Indeed for $x \neq 1$ we have $$\sum_{n=0}^\infty x^n(1-x) = \sum_{n=0}^\infty x^n - \sum_{n=1}^\infty x^{n} = 1,$$ but for $x = 1$ the series is $0$. If the convergence was uniform, the resulting function would be continuous, which it isn't.

There is however a result called Weierstrass M-test. It says that if each of the functions $f_n$ is bounded, i.e. $\sup_{x \in [0,1]} |f_n(x)| \le M_n$ for some constant $M_n > 0$, and if $\sum_{n=0}^\infty M_n < \infty$, then the convergence is uniform. (Here I have taken $[0,1]$ as the domain of the functions. You may replace this by any set.)

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This example is neither absolutely nor uniformly convergent. I say that it is not absolutely convergent coz this series has 1 as convergence on (0,1] while 0 on point 0 which means it is divergent on [0,1]. –  Frank_W Dec 19 at 15:57

In view of the reference to Dini one might ask, what if all $f_n$ and $f$ are continuous functions on a compact set $K$? Thus if all $f_n \ge 0$, Dini would say the convergence is uniform on $K$. But without that hypothesis it's not true. Consider $K=[0,1]$ where
$f_{2n-1}$ is the piecewise-linear interpolation of $f_{2n-1}(0) = 0$, $f_{2n-1}(1/(2n)) = 1$, $f_{2n-1}(1/n) = 0$, $f_{2n-1}(1) = 0$, while $f_{2n}(x) = - f_{2n-1}(x)$. Then $\sum_{n=1}^\infty f_n(x)$ converges absolutely to $f(x) = 0$ but the odd partial sums have maximum value 1.

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