$(a_n)$ a sequence of continuous function $a_n:\mathbb{R}\rightarrow\mathbb{R}$. Non-increasing, non-negative. If $\sum a_n(x)$ convergers for all $x$ then the function $g(x)=\sum a_n(x)$ is continuous.

EDIT: Call $g_n(x)=\sum_{k=0}^n a_n(x)$, this is continuous and $g_n(x)\rightarrow g(x)$.

$|g(x)-g(x_0)|\leq|g(x)-g_n(x)|+|g_n(x)-g_n(x_0)|+|g_n(x_0)-g(x_0)|$. The second addend is less than $\varepsilon$ if $|x-x_0|\leq\delta$, I want to estimate the first and third addend. Now

$|g(x)-g_n(x)|\leq\sum_{k=n+1}^\infty a_k(x)$ and because the series is convergent this is less that $\varepsilon$ if $n>n_1$. The same for the third addend if $n>n_2$. So if we take $N=\mathrm{max}\{{n_1,n_2}\}$ then we have the thesis. There sould be something wrong because I'm not using the hypothesis that the $a_n$ are non-increasing, could you help me to find the mistake?

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    $\begingroup$ Look for a counterexample. $\endgroup$ – GEdgar Oct 16 '11 at 23:25
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    $\begingroup$ To help in carrying out GEdgar's hint: Where do we look to find counter examples? Your sequence of partial sums $ \sum_{k=0}^n a_k (x) $ is continuous, and it converges pointwise to $ \sum_{k=0}^{\infty} a_k(x) $. If the convergence was not just pointwise, but uniform, then there is a theorem that the limit function would also be continuous. Go through the proof of that statement, look carefully at the details to see where uniform convergence is required and why pointwise convergence is not enough - then construct a function to exploit that. $\endgroup$ – Ragib Zaman Oct 17 '11 at 1:15
  • $\begingroup$ I think the point of this exercise is to show that if $(a_n(x))$ is a non-negative, non-increasing sequence of continuous functions, pointwise and uniform convergence of its sum are equivalent. Or disprove it. $\endgroup$ – Arthur Oct 17 '11 at 2:40
  • $\begingroup$ I've tried to follow your hints, I edited the question. But there should be something wrong in my answer, could you help me please? $\endgroup$ – Mec Oct 18 '11 at 1:14

Nate has already pointed out the problem with your proof. Regarding a fix: Are you familiar with the Weierstrass $M$-test? One approach would be to use this, noting that $a_n(t)$ is majorized by $a_n(x)$ whenever $t\in[x,\infty)$. This will give you uniform convergence (hence continuity) on every interval $[x,\infty)$. (The proof of the Weierstrass test is extremely simple, and is given on the Wikipedia page I linked.)

Another approach would be Dini's theorem (again, very simple proof given on the Wikipedia page) on compact sets, using the fact that the difference $g-g_n=\sum_{k=n+1}^{\infty}a_n(x)$ is non-increasing in both $n$ and $x$.

As a side note, both of these theorems are very easily proved (i.e., no more difficult than solving this problem) and the ideas involved in the proofs come up very often in questions such as this one.


Be careful with your quantifiers. Your $\delta$ depends on $n$, and then you change $n$ depending on $x$.

I suggest you write out your (corrected) proof very carefully, being explicit about all the choices you make (e.g. how exactly do you choose $\delta$, and how can you prove that this $\delta$ does what you want?).

You didn't use the assumption that the $a_n$ are nonnegative either; both are necessary, as Nick Strehlke pointed out.

  • $\begingroup$ I think that the non-increasing assumption might be necessary. For example, let $a_n(x)\geq0$ be continuous and supported in the open interval $(1/(n+1),1/n)$, with maximum value $1$. (Correct me if I'm wrong :)) $\endgroup$ – Nick Strehlke Oct 18 '11 at 5:33
  • $\begingroup$ @NickStrehlke: You're right. I removed my false statement. I misremembered Dini's theorem. $\endgroup$ – Nate Eldredge Oct 18 '11 at 13:49

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