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!

  • $\begingroup$ What is the difference between $S_N$ and $\phi_N$ above, their definitions are the same... $\endgroup$
    – copper.hat
    Jun 4 '12 at 17:31
  • $\begingroup$ 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. $\endgroup$
    – user20266
    Jun 4 '12 at 17:33
  • $\begingroup$ @copper.hat $S_N$ is from 0 to $N-1$ $\endgroup$
    – user32778
    Jun 4 '12 at 17:44
  • $\begingroup$ @Thomas right, but we're interested in a more specific proof as stated above. $\endgroup$
    – user32778
    Jun 4 '12 at 17:46
  • 1
    $\begingroup$ 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. $\endgroup$
    – leo
    Jun 4 '12 at 17:53

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.


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