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I am almost successful in this proof but my argument needs some touches. It suffices to show that the sequence of partial sums, being continuous, converges uniformly, Then, the limit function must be continuous. Here is the question

If $\sum^{\infty}_{k=0}a_k$ converges, then $\sum^{\infty}_{k=0}a_k x^k$ converges to a continuous function on $(-1,1)$

Let $\epsilon>0$ be given. Convergence of $\sum^{\infty}_{k=0}a_k$, implies that there exists $N$ such that

$$\left| \sum^{m}_{k=n+1}a_k \right|=\left| \sum^{m}_{k=0}a_k-\sum^{n}_{k=0}a_k \right|<\epsilon,\;\;\forall\;m\geq n\geq N.$$ Define $$A_n=\sum^{n}_{k=0}a_k $$ Since $\{A_n\}_n$ converges then it is bounded. So, there exists $M>0$ such that $$\left|A_n\right|\leq M,\;\;\forall\;n\geq 1. $$ Convergence of $\{\left|x\right|^m\}_m$ to zero implies that there exists $N_0$ such that $$\left|x\right|^m\leq \frac{\epsilon}{M},\;\;\forall\;n\geq N_0. $$ Let $m,n\in \Bbb{N}$ such that $m\geq n$ and $\bar{N}=\max\{N_0,N \}$. Then,

\begin{align}\left|\sum^{n}_{k=n+1}a_k x^k \right|&=\left|\sum^{m-1}_{k=n+1}A_k (x^k-x^{k+1}) -A_m x^m\right|\\\leq &\left|\sum^{m-1}_{k=n+1}A_k (x^k-x^{k+1}) -A_m x^m\right|\\\leq &\sum^{m-1}_{k=n+1}\left(\left|A_k\right| \left|x^k-x^{k+1}\right|\right) +\left|A_m\right| \left|x\right|^m\\<&\epsilon\sum^{m-1}_{k=n+1}\left|x^k-x^{k+1}\right|+M\left(\frac{\epsilon}{M}\right)\end{align} My problem now, is how to bound $$\sum^{m-1}_{k=n+1}\left|x^k-x^{k+1}\right|.$$ The sequence $\{ x^k\}_k$ is decreasing only in $[0,1)$ and not in $(-1,0)$. So, I cannot have $$\sum^{m-1}_{k=n+1}\left|x^k-x^{k+1}\right|=\sum^{m-1}_{k=n+1}\left(x^k-x^{k+1}\right).$$ If this were possible, the proof is done. Is there any way I can bound this? Any help will be appreciated.

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Let $0<\epsilon<1.$ Now, $\sup |a_k|<M<\infty$ since $\sum a_k$ converges, and ths implies that $|a_k|\cdot | x(1-\epsilon)|^k<M$ for all $x\in (-1,1)$ so by the Weierstrass $M$-test, $\sum a_k x^k(1-\epsilon)^k$ converges uniformly--hence to a continuous function--on $(-1,1)$. But now since $\epsilon$ is arbitrary, the result follows.

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  • $\begingroup$ Sorry, how did you get $|a_k|\cdot | x(1-\epsilon)|^k<M$ for all $x\in (-1,1)$? $\endgroup$ – Micheal Jan 28 at 21:01
  • $\begingroup$ because if $x\in (-1,1)$ and $0<\epsilon<1$ then $|a_k|\cdot | x(1-\epsilon)|^k<|a_k|\cdot 1<M.$ $\endgroup$ – Matematleta Jan 28 at 21:05
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    $\begingroup$ I like very much your solution (+1) ! $\endgroup$ – Duchamp Gérard H. E. Jan 29 at 8:25
  • $\begingroup$ @DuchampGérardH.E. Thanks! $\endgroup$ – Matematleta Jan 29 at 15:21
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I give you an alternative proof, much more simpler. From Cauchy's convergence criterion: $$\sum_{n=k}^\infty a_n \text{ converges} \implies \limsup_{n\to \infty} \sqrt[n]{|a_n|} \leq 1 $$ then $$\limsup_{n\to\infty} \sqrt[n]{|a_nx^n|} = |x|\cdot \limsup_{n\to\infty}\sqrt[n]{|a_n|} <1 \implies \sum_{n=k}^\infty a_nx^n\text{ converges} $$

One way of proving that $\sum_{n=k}^\infty a_nx^n$ is continuous would be by Lebesgue's dominated convergence theorem. For any $x\in (-1, 1)$ there exist a closed interval $[a, b]\subset (-1, 1)$, such that $x$ belongs to it's interior. Then $$ |a_nx^n|\leq \max\{|a_na^n|, |a_nb^n|\} $$ and continuity follows by applying Lebesgue's theorem.

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  • $\begingroup$ Is there any other way you can go about it? I don't want to use Lebesgue as it comes after this topic. $\endgroup$ – Micheal Jan 28 at 20:48
  • $\begingroup$ @Micheal The topic about Lebesgue integrals doesn't depend on this theorem and is very useful. I suggest you read it. Anyway, I wanted this more to be a cool application of this theorem. $\endgroup$ – Jakobian Jan 28 at 20:52
  • $\begingroup$ Lebesgue's theorem on what space, $\ell^\infty(\mathbb N)$? How does this get you continuity? I feel like the correct application of dominated convergence requires something like a dominating function in $\ell^1(\mathbb N ; C([a,b]))$ (which is a messy rephrasing of Weierstrass M-test) $\endgroup$ – Calvin Khor Jan 28 at 21:33
  • $\begingroup$ @CalvinKhor Heine's definition of continuity. We just take counting measure. Not sure why would you complicate it even more and introduce $l^p$ spaces. It's already simple enough. And no, I don't think it's messy, I think it's quite understandable. $\endgroup$ – Jakobian Jan 28 at 22:25
  • $\begingroup$ I didn't say yours is messy, I said what I said is messy, because its the only way I see to apply DCT to obtain the result (It is actually just a reformulation of the standard proof, Weierstrass M-test). Sorry for the confusion. That said I do not see how the convergence of the integral $\sum_{n=0}^\infty a_n x^n$ for every $x$ (which is the conclusion of DCT) would imply the continuity in $x$? $\endgroup$ – Calvin Khor Jan 28 at 22:37
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Note that you do not need uniform continuity on the whole interval (-1,1) - it is enough if you show uniform continuity on (-r,r), for all r in (0,1); the hint about uniform boundedness of $|a_n|$ should allow you to deduce said uniform continuity on (-r,r) by comparison with the geometric series.

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  • $\begingroup$ Exact ! I This hint about uniform boundedness is crucial. $\endgroup$ – Duchamp Gérard H. E. Jan 28 at 21:06
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A weaker hypothesis allows to see better. I propose you a lemma in Dieudonné ``Infinitesimal Calculus'' here

If $a_n$ is uniformly bounded i.e. it exists $B$ such that $|a_n|\leq B$ then $\sum_{n=0}^{+\infty}a_nx^n$ converges to a continuous function in $(-1,1)$.

As pointed by Conrad, it suffices to prove uniform continuity on each $[-r,r]$ for all $0\leq r<1$. For each of these $r$ and $f:(-1,1)\to \mathbb{R}$, note $||f||_r=sup_{s\in [-r,r]}|f(s)|$. Then, if $a_n$ is uniformly bounded by $B$, one has $$ \sum_{n=0}^{+\infty}||a_nx^n||_r\leq B\sum_{n=0}^{+\infty}r^n=\frac{B}{1-r}<+\infty $$ which proves uniform continuity on each on each $[-r,r]$ and then continuity on $(-1,1)$.

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  • $\begingroup$ Please, how do I have access to the book? $\endgroup$ – Micheal Jan 28 at 20:41
  • $\begingroup$ I added the reference. $\endgroup$ – Duchamp Gérard H. E. Jan 28 at 20:55

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