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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.

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  • $\begingroup$ Why is it uniform on (-1,1)? I think it is only uniform on closed subsets of (-1,1) $\endgroup$ – Calvin Khor Feb 5 '18 at 10:23
  • $\begingroup$ Yes. But [-1+$\epsilon$,1-$\epsilon$] for any $\epsilon>0$ is just (-1,1). $\endgroup$ – Math The Novice Feb 5 '18 at 10:27
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    $\begingroup$ $x^n$ converges locally uniformly to 0 on (0,1) but not uniformly $\endgroup$ – Calvin Khor Feb 5 '18 at 10:28
  • $\begingroup$ @CalvinKhor Excuse me, what does this imply? I have no exposure to the term ''locally uniform''. $\endgroup$ – Math The Novice Feb 5 '18 at 10:34
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    $\begingroup$ locally uniform is what i said above: uniform on closed and bounded subsets $\endgroup$ – Calvin Khor Feb 5 '18 at 10:35
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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.

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  • $\begingroup$ Sorry if I miss something. Isn't "the $\sum_{n=0}^{\infty} c_n$ exists" in the hypothesis of the theorem? $\endgroup$ – Math The Novice Feb 5 '18 at 11:09
  • $\begingroup$ @MathTheNovice but you use the statement i quoted above as a black box, and its not true. $\endgroup$ – Calvin Khor Feb 5 '18 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? $\endgroup$ – Math The Novice Feb 5 '18 at 12:56
  • $\begingroup$ @Math No. The $c_n $ don't need to be positive. $\endgroup$ – Andrés E. Caicedo Feb 5 '18 at 13:07
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    $\begingroup$ @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. $\endgroup$ – Andrés E. Caicedo Feb 5 '18 at 13:58

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