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I want to prove that every power series is continuous but I am stuck at one point.

Let $\sum\limits_{n=0}^\infty a_n(x-x_0)^n$ a power series with a radius of convergence $r>0$ and let $D:=\{x\in\mathbb R:|x-x_0|<r\}$. Then $S(x)=\sum\limits_{n=0}^\infty a_n(x-x_0)^n$ is continuous on $D$.

Proof: Let $x\in D$ and $r_0\in\mathbb R^+$ such that $|x-y_0|<r_0<r$ and for any positive integer $N$ let $S_N(y)=\sum\limits_{n=0}^{N-1} a_n(y-x_0)^n$ and $\phi_N(y)=\sum\limits_{n=N}^\infty a_n(y-x_0)^n$.

Since the power series converges uniformly on the closed disk $|y-x_0|\leq r_0$, we may choose a positive integer $N_\varepsilon$ such that $|\phi_{N_\varepsilon}|<\frac\varepsilon3$ for all $|y-x_0|\leq r_0$.

Since $S_{N_\varepsilon}$ is polynomial we can choose a $\bar\delta>0$ such that $|S_{N_\varepsilon}(y)-S_{N_\varepsilon}(x)|<\frac\varepsilon3$. So we get $|S(y)-S(x)|<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon$ for $|y-x|<\delta$

Why is in the violet term $|\phi_{N_\varepsilon}|<\frac\varepsilon3$ correct? Don't I have two terms in the absolute value? Thanks for helping!

  • What is the difference between $S_N$ and $\phi_N$ above, their definitions are the same... – copper.hat Jun 04 '12 at 17:31
  • the uniform limit of continuous functions is continuous. If you already do know that power series converge uniformly there does not remain anything to prove. –  Jun 04 '12 at 17:33
  • @copper.hat $S_N$ is from 0 to $N-1$ –  Jun 04 '12 at 17:44
  • @Thomas right, but we're interested in a more specific proof as stated above. –  Jun 04 '12 at 17:46
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    If you want to prove that $S(x)$ is continuous in $D$, given $c\in D$, and given $\epsilon\gt 0$ you need to find a $\delta\gt 0$ such that $$|x-c|\lt\delta\implies |S(x)-S(y)|\lt\epsilon.$$ Is that what you are trying to say? Once you understand that, the proof is as you wrote, by dividing the series $S(x)-S(c)$ in two parts. – leo Jun 04 '12 at 17:53
  • I think you want $$\phi_N(y)=\sum_{n=N}^\infty a_n(y-x_0)$$ – leo Jun 04 '12 at 17:56
  • @niro Did you notice all the powers are missing from $,(x-x_0),$? These are not power series but only series. – DonAntonio Jun 04 '12 at 18:10
  • I guess it would be $(x-x_0)^n$. Please consider review your post. If you put effort on your question it is easier to help you :-) – leo Jun 04 '12 at 18:26
  • @leo EXACTLY, this is what I want to do! At the end I have $|S(y)-S(x)|=|S_{N_\varepsilon}(y)+\phi_{N_\varepsilon}(y)-S_{N_\varepsilon}(x)-\phi_{N_\varepsilon}(x)|$ and then it's via triangle inequality as stated above. So why is $|\phi_{N_\varepsilon}(y)|<\frac\varepsilon3$ as above correct this way? –  Jun 04 '12 at 18:36
  • I prefer to use the Weierstrass M-test and completeness of continuous functions under the $\sup$ norm. – copper.hat Jun 04 '12 at 19:39
  • @copper.hat post an answer, it would be useful! – leo Jun 04 '12 at 19:41

1 Answers1

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Choose $z \in D$ and select $\delta >0$ so that $C = \overline{B}(z,\delta) \subset D$. Let $\rho = |z-x_0|+\delta$, note that $\rho < r_0$ and that $C \subset \overline{B}(x_0,\rho) \subset D$.

Then let $M_n = |a_n|\rho^n$, and note that $|a_n(x-x_0)^n | \leq M_n$, $\forall x \in C$, and $\sum M_n < \infty$.

Hence we can use the Weierstrass M-test to conclude that the series $\sum a_n (x-x_0)^n$ converges uniformly on $C$. Since each of the functions $\sum_{n<N} a_n (x-x_0)^n$ is trivially continuous, it follows from the uniform limit theorem that the limit function $x \mapsto \sum a_n (x-x_0)^n$ is continuous on $C$. Since $z\in D$ was arbitrary, the desired result follows.

copper.hat
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  • Why did you introduce $\rho$? Can't we simply argue like: $$|a_n(x-x_0)^n|\leq |a_n||(z+\delta)-x_0|^n=:M_n$$ for all $x\in C$. And $\sum M_n<\infty$ follows because $z+\delta\in D$ – Philipp Feb 03 '25 at 22:29
  • @Philipp To show that it is continuous, I need some ball around $z$ that is contained in $D$ for which the summation is defined. – copper.hat Feb 03 '25 at 23:18
  • I didn't omit the ball, I still use $C = \overline{B}(z,\delta)$. I have only suggested that the other ball $ \overline{B}(x_0,\rho)$ seems not necessary when finding an upper bound for the M-test. – Philipp Feb 03 '25 at 23:38
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    @Philipp I don't really follow what you are asking. For example, you seem to imply that $|x-x_0| \le |z+\delta-x_0|$ but that is not necessarily true. We must have by assumption that $\delta < r-|z-x_0|$, or equivalently, $\delta+|z-x_0| < r$. For convenience, I let $\rho = \delta+|z-x_0|$. – copper.hat Feb 04 '25 at 04:33
  • Note that $\sum_n |a_n| r^n$ is not necessarily summable, but for any $ r' \in [0,r)$, $\sum_n |a_n| (r')^n$ is summable. – copper.hat Feb 04 '25 at 04:35
  • How do you know that $\sum M_n<\infty$? Maybe it's obvious to you? But to make it more understandable I would suggest to add another intermediate step to complete the argument. The power series only converges on $]x_0-r,x_0+r[$ if we consider the $a_n$ without the absolute value bars. (I ran into that problem in a related question see https://math.stackexchange.com/q/5031562/579544 ) – Philipp Feb 04 '25 at 23:12
  • That is beyond the scope of the question. I can't answer without knowing how you define the radius of convergence in the first place, the definition I use is https://en.wikipedia.org/wiki/Radius_of_convergence#Theoretical_radius and it is a theorem that it converges absolutely inside (strictly) the ROC and diverges outside (strictly). – copper.hat Feb 05 '25 at 00:30