Refering to Theorem 8.2 in Baby Rudin

8.2 Theorem Suppse $\sum c_n$ converges. Put $$f(x)=\sum_{n=0}^{\infty} c_nx^n \ \ \ \ (-1<x<1)$$ Then $$\lim_{x\rightarrow1}f(x)=\sum_{n=0}^{\infty}c_n$$

The proof in Rudin is that outlined in Wikipedia

However, Factoring out $(1-x)$ seems not natural for me.

The theorem looks like a extension from Theorem 8.1.

Indeed, by imitating the proof of Rudin 7.11 $$|f(t)-\sum^{\infty}_{n=0}c_n|\leq|\sum_{n=0}^{\infty}c_n-\sum_{n=0}^{N}c_n|+|\sum_{n=0}^{N}c_n-\sum_{n=0}^{N}c_nt^n|+|\sum_{n=0}^{N}c_nt^n-f(t)|$$ where $t \in(-1,1)$

1.The first term can be arbitrary small for large N since $\sum c_n$ converges.

2.The second term can be small when $t\rightarrow1$ since polynomial is continuous.

3.Since f(x) is uniform convergent on $(-1,1)$, the third term can also be arbitrary small.

Then, I can also conclude the same result as in Theorem 8.2.

Could anyone kindly explain why Rudin uses a different approach (like factoring out (1-x)) or my reasoning has flaws, if there is any ?

Summary of the discussion :

The problem in my proof arises from the third term in the inequality, where I mistake the order to take limits (I implicitly make $t \rightarrow 1$ and then $ N \rightarrow \infty$. This is wrong).

The motivation (I guess) in Rudin's proof is from Rudin Theorem 3.42, where we study the criteria to test conditional convergent series. Theorem 8.2 has a similar situation, the sum $\sum c_n$ may very well be not absolutely convergent.

  • $\begingroup$ Why is it uniform on (-1,1)? I think it is only uniform on closed subsets of (-1,1) $\endgroup$ Feb 5, 2018 at 10:23
  • $\begingroup$ Yes. But [-1+$\epsilon$,1-$\epsilon$] for any $\epsilon>0$ is just (-1,1). $\endgroup$ Feb 5, 2018 at 10:27
  • 3
    $\begingroup$ $x^n$ converges locally uniformly to 0 on (0,1) but not uniformly $\endgroup$ Feb 5, 2018 at 10:28
  • $\begingroup$ @CalvinKhor Excuse me, what does this imply? I have no exposure to the term ''locally uniform''. $\endgroup$ Feb 5, 2018 at 10:34
  • 1
    $\begingroup$ locally uniform is what i said above: uniform on closed and bounded subsets $\endgroup$ Feb 5, 2018 at 10:35

1 Answer 1


You claim that $$\left | \sum_{n=0}^N c_n x^n - f(x) \right| \xrightarrow[N\to\infty]{} 0 \quad \text{uniformly in }(-1,1) $$ this is not necessarily true. For it to have any chance to be true, it must make reference to the extra assumption that $\sum_{n=0}^\infty c_n$ exists, because its not true for arbitrary power series that converge (locally uniformly) on $(-1,1)$.

For instance, consider $f(x) = \sum_{n=0}^\infty x^n = \frac1{1-x}$. Its partial sums cannot converge uniformly on any $(1-\epsilon,1)$ since the partial sums are bounded but the limit is unbounded.

  • $\begingroup$ Sorry if I miss something. Isn't "the $\sum_{n=0}^{\infty} c_n$ exists" in the hypothesis of the theorem? $\endgroup$ Feb 5, 2018 at 11:09
  • $\begingroup$ @MathTheNovice but you use the statement i quoted above as a black box, and its not true. $\endgroup$ Feb 5, 2018 at 11:11
  • $\begingroup$ So basically, if $\sum^{\infty}_{n=0} c_n $ exists, apply M-test and my statement can be proven true. Yes? $\endgroup$ Feb 5, 2018 at 12:56
  • $\begingroup$ @Math No. The $c_n $ don't need to be positive. $\endgroup$ Feb 5, 2018 at 13:07
  • 1
    $\begingroup$ @Math No. $f (x) $ is absolutely convergent in the interior of its interval of convergence. This does not make the $c_n $ be positive, so the M-test may not apply. But yes, if you know to begin with that $\sum c_n $ converges absolutely, then you can proceed as you suggest. But this case is significantly easier than the general case, where $\sum c_n $ only converges conditionally. $\endgroup$ Feb 5, 2018 at 13:58

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.