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Theorem 9.7 Assume that $\sum f_n(x)= f(x)$ (uniformly on $S$). If each $f_n$ is continuous at a point $x_0$ of $S$, then $f$ is also continuous at $x_0$.

Proof. I define $s_n(x) = \sum_{k=1}^n f_k(x)$. I know that the sum of continuous functions is a continuous function. For this reason $s_n(x)$ is continuous at $x_0$ in $S$.In addition I know that if each $f_k$ in $\{f_k\}$ is continuous, then also the limit function of $\{f_k\}$, $f$ will be continuous. $\{f_k\}$ converges uniformly to $f$.

We know that $\sum f_n(x)$ converges. For $n>N,\;|f(x)-\sum_{k=0}^nf_k(x)|<\epsilon$. To make the last step in this proof I say that for $n>N$

$$\sum_{k=n+1}^\infty f_k(x)= f(x)-\sum_{k=0}^nf_k(x)\\\sum_{k=n+1}^\infty f_k(x)+\sum_{k=0}^nf_k(x)=f(x)$$

Being the sum of two continuous functions $f(x)$ is also continuous and I hope I proved the theorem. I mentioned all the hints given by Apostol. Could you please check my proof? Is it correct?

Thank you in advance.

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3 Answers 3

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Theorem: If $(f_n)$ is a sequence of functions uniformly convergent to $f$ and every function $f_n$ is continuous at $x_0$ then $f$ is continuous at $x_0$.

Proof: Using the definition of the uniform convergence and the continuity of $f_n$ and the fact $$|f(x)-f(x_0)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|$$ we find easily the resut.

Now apply the previous theorem to the function $$s_n=\sum_{k=0}^nf_k$$

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  • $\begingroup$ $+1{}{}{}{}{}{}{}$ $\endgroup$
    – Mikasa
    Commented Jan 3, 2014 at 17:58
  • $\begingroup$ @Sami I tried to follow your advice in a new proof. $\endgroup$
    – Charlie
    Commented Jan 3, 2014 at 18:43
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How do you know that $\sum_{k = n + 1}^{\infty} f_k(x)$ is continuous? An infinite sum of continuous functions does not need to be continuous; it's not entirely clear how you're proving this. I think what you're trying to say is this:

If a sequence of continuous functions converges uniformly, the limit function is continuous.

If so, then you're pretty much done already: The sequence $s_n$ converges to $f$ uniformly, and we're done. You've written that $f_k$ converges uniformly to $f$, which isn't true.

Also, the equalities you've written relating the two parts of the infinite series are true for every $n$, and don't at all relate to the $\epsilon$ you've chosen.

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Theorem 9.7 Assume that $\sum f_n(x)= f(x)$ (uniformly on $S$). If each $f_n$ is continuous at a point $c$ of $S$, then $f$ is also continuous at $c$.

Proof. $\sum f_n(x)$ converges uniformly to $f(x)$ in $S$.

  1. For the definition of uniform convergence $\exists\;\epsilon>0\;\text{and an}\;n>N \colon|\sum_{k=0}^n f_n(x)-f(x)|<\frac \epsilon{3}$
  2. For continuity of $s_n(x)=\sum_{k=0}^nf_k(x)$ at $c$, $\exists\;B(c)\colon x\in B(c)\cap S $ $$|s_n(x)-s_n(c)|<\frac \epsilon{3} $$
  3. But

$$|f(x)-s_n(c)|<|f(x)-s_n(x)|+|s_n(x)-s_n(c)|+|s_n(c)-f(c)|$$

and each of the terms on the right side of the last inequality is lower than $\frac \epsilon{3}$. Finally, If $x\in B(c)\cap S$ we have $$|f(x)-s_n(c)|<\epsilon.$$

This is my attempt to prove this theorem using your precious advices. Thank you.

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