Whether in Real Analysis or by Open Set Def of Continuity in Topology, it is easy to show that the sum and product of a FINITE number of continuous functions are also continuous functions. That is, assuming that $f_1, ..., f_m:\Bbb R\rightarrow\Bbb R$ are continuous, then $S:\Bbb R\rightarrow\Bbb R$ and $P:\Bbb R\rightarrow\Bbb R$, defined by $S(x) = f_1(x) + ... + f_m(x)$ and $P(x) = f_1(x) \times ... \times f_m(x)$, are continuous.

But many analytic functions that are continuous can be written in their expanded form (by Taylor Series), which are the sum and product of INFINITE functions. My question is, EVEN if there is another way to show that some/all analytic functions are continuous (which I don't know that way), still we should prove from "the sum and product of infinite functions" way.

Would you please help me regarding the question? I think one of Topology or Analysis way of proof should be enough, because, as we can prove, the topological definition of continuity is equivalent to the $\epsilon - \delta$ definition for functions that map $\Bbb R$ to $\Bbb R$.

EDIT: Let me rephrase it: limit of sum of two functions exists if limit of each of the two functions exists. If sum of in finite number of functions is a function that has limit in some point, is it mean that we are allowed to say that for this type of function, sum of limit infinite number of functions exists since limit of sum of those infinite number of functions exists?

  • $\begingroup$ The statement in your title isn't true. You should be able to find functions $f_n$ all continuous whose limit $f$ is not continuous - continuity is not a property which is guaranteed to be preserved when taking limits. Set $g_1=f_1$ and otherwise $g_n=f_n-f_{n-1}$. The $g_n$ are continuous and their sum tends to the discontinuous limit $f$. $\endgroup$ – Mark Bennet Jan 3 '15 at 14:24
  • $\begingroup$ If $\{f_n\}$, a sequence of continuous function, converges uniformly to $f$, the result is true. $\endgroup$ – jdoicj Jan 3 '15 at 14:27

Consider the function defined by $$\begin{eqnarray}f(x) &= &\begin{cases} 1 &\mbox{if}\quad 0<x<\pi\\ 0 &\mbox{if}\quad x=0\\ -1 &\mbox{if}\quad -\pi<x<0 \end{cases}\\ f(x+2n\pi) &=& f(x) \quad \mbox{if}\quad n \in \mathbb N \end{eqnarray}.$$ This is an example of a "square wave." It is not continuous, but Fourier analysis gives an infinite sum of functions (all of the form $a \sin(k\pi)$) that add up to this function.

Regarding your edited question, on the other hand, the only way I know of to even define the sum of an infinite number of functions is that the sum is the limit function $f$ of the sequence of finite function sums $f_n$ as $n$ goes to infinity, if such a function $f$ exists. If we know that $f$ has a limit at a certain point, that is, $\lim_{x\to\infty} f(x)$ exists for a certain value of $x$, then $f$ is continuous at $x$, and so is the infinite sum of functions (since that sum is defined as $f$).

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    $\begingroup$ The square wave is a fantastic counterexample! Thank you very much. $\endgroup$ – L.G. Jan 3 '15 at 14:54
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    $\begingroup$ So in summary, we can say: the sum of a infinite number of continuous functions may or may not be continuous function. In fact, it depends on the result of the sum. Also, the limit of sum of infinite number of continuous functions may or may not be equal to the sum of limits of those continuous functions. Am I right? $\endgroup$ – L.G. Jan 3 '15 at 14:58
  • $\begingroup$ @David.K.: Regarding the Uniform Convergence Theorem (in your last paragraph), what if we define $f_n=\sum\limits_{i=1}^n a \sin(n\pi)$ instead of $g_n=a \sin(n\pi)$? $f_n$ is a sequence of continuous functions that converges uniformly to $f :$"square wave" , but $f :$"square wave" is not continuous. $\endgroup$ – L.G. Jan 4 '15 at 4:42
  • $\begingroup$ @Ali.E. The last paragraph is really a different case, which I added only because you added a paragraph to the question that seemed to be asking something a bit different than the main question. I've edited that last paragraph in an attempt to clarify what it says. $\endgroup$ – David K Jan 4 '15 at 19:00
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    $\begingroup$ @Ali.E. Regarding the questions in comments: "the sum of a infinite number of continuous functions may or may not be continuous function" is true. "The limit of sum of infinite number of continuous functions may or may not be equal to the sum of limits of those continuous functions" is also true, because the sum may not have a limit at that point. Finally, mere convergence to a function is not necessarily uniform convergence, and the convergence to a square wave is definitely not uniform. If you wanted only uniform convergence, that is a different question. $\endgroup$ – David K Jan 4 '15 at 19:07

I might be wrong here, but I do not think your assumption is right. If we make an infinite series of functions (Each term of the series would correspond to a term in the infinite set of continuous functions), and assume that this series converges to a function $f$, then we cannot conclude that the $f$ is continuous unless that convergence was uniform convergence.

I will quote an example from Apostol's Calculus. Let $f_n(x)=x^n$ where $0\leq x\leq1$. $f(x)=0$ when $0\leq x <1$, $f(x)=1$ when $x=1$. Thus, although the sequence of functions converges, since the convergence isn't uniform, the limit function is non-continuous despite the functions in the sequence being continuous.

I cannot come up with a series like that now, but since a series is basically a sequence of finite sums, you can apply the same logic to it. I hope this has helped.

  • $\begingroup$ Just make a telescopic series out of the sequence: the series with terms $a_0,a_1-a_0,a_2-a_1,\dotsc$ has the same character as the sequence you started with. $\endgroup$ – egreg Jan 3 '15 at 15:05

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