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

Question

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.

Answer 1

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.

Answer 2

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 $(-1,1)$.

Answer 3

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.

Answer 4

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.