Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having some difficulty with the following question:

Let $(f_{n}(x))$ be a uniformly convergent functions sequence in $(a,b)$ (where b can be $\infty$) such that $(f_{n}(x)) \to f(x)$. Suppose that for almost all $n$, the limit $\displaystyle \lim_{x\to b^{-}}f_{n}(x)$ exists.

Prove that $\displaystyle \lim_{x\to b^{-}}f(x)$ exists and $\displaystyle \lim_{n\to \infty}\left (\lim_{x\to b^{-}}f_{n}(x)\right )=\lim_{x\to b^{-}}f(x)$

I tried a method that I see often in theorems regarding uniform convergence:

First, let $\displaystyle \left ( a_{n}=\lim_{x\to b^{-}}f_{n}(x) \right )_{n\geq N}$. Such an $N$ exists.

We want to show that $\forall \epsilon,\ \exists \delta,\ \forall x \in (b-\delta, b),\ |f(x)-L|\lt \epsilon$.

We can write $|f(x)-L|=|f(x)-f_{n}(x)+f_{n}(x)-a_{n}+a_{n}-L|$, then:

$$|f(x)-L|\lt\overbrace{|f(x)-f_{n}(x)|}^{\lt \epsilon / 3}+\overbrace{|f_{n}(x)-a_{n}|}^{\lt \epsilon \lt 3}+\overbrace{|a_{n}-L|}^{\text{?}}$$

But I don't know how to deal with $|a_{n}-L|$. Had $a_{n}\to L$, (which is in fact the second part of the question) then I could make sure it's no more than $\epsilon / 3$, but I don't know if $a_{n}$ converges yet and if it converges to $L$.

Note: couldn't think of a better title, if anyone does, feel free to modify it.

share|cite|improve this question
What is $L$? Is it $\lim\limits_{x \to b^{-}} f(x)$? – PEV Jan 17 '11 at 20:49
@Trevor: Yes, it is supposed to be. One has to show that this exists, and that $L$ is also $\lim_{n\to\infty}a_n$. – Jonas Meyer Jan 17 '11 at 20:52
up vote 2 down vote accepted

Right, the trouble is that you don't have an expression up front for $L$, so you're going to have a hard time showing that $f(x)$ converges to it (as $x \to b^-$). A good approach in this case is to use completeness of $\mathbb{R}$; then all you have to do is show that $f(x)$ is Cauchy as $x \to b^-$, and it follows that it converges to something, which you can then call $L$.

So try to show the following: for all $\epsilon > 0$, there exists $\delta > 0$ such that for all $x,y \in (b, b+\delta)$ we have $|f(x) - f(y)| < \epsilon$. By completeness, it will follow that $\lim_{x \to b^-} f(x)$ exists, and you can call that number $L$. (If you think of completeness in terms of Cauchy sequences, then note that the statement implies that for any sequence $x_n \downarrow b$, we have that $f(x_n)$ is a Cauchy sequence, hence convergent to some number $L$, and that $L$ is the same for all such sequences $x_n$.) It should not be hard then to show that $a_n \to L$ as well.

share|cite|improve this answer
Thanks. Actually I have a theorem in my textbook that every Cauchy sequence is convergent so I can use that. On how to show $f(x)$ is Cauchy, would the following work? $$|f(x)-f(y)|\leq |f(x)-f_{n}(x)|+|f_{n}(x)-f_{m}(y)|+|f_{m}(y)-f(y)|$$ Using uniform convergence of $f_{n}(x)$, the first and third terms are less than $\epsilon/3$, and the second is also less than $\epsilon/3$ since $f_{n}(x)$ is also Cauchy. – daniel.jackson Jan 17 '11 at 21:15
I would actually just take $n=m$ in your inequality. For the second term you have to use the fact that $\lim_{x \to b^-} f_n(x)$ exists for each $n$ (so think about splitting it into two terms). – Nate Eldredge Jan 17 '11 at 22:20
I thought I knew how to show that $a_{n}\to L$ but now I'm not so sure. We know that $\lim_{x\to b^{-}} f(x)=L$, and now I'm trying to look at $$|a_{n}-L|\leq |a_{n}-f(x)|+\overbrace{|f(x)-L|}^{\lt \epsilon / 2}$$ but I'm not sure what to make of $|a_{n}-f(x)|=|\lim_{x\to b^{-}}f_{n}(x)-f(x)|$. I'm really finding it confusing having to deal with both $n$ and $x$ at the same time. – daniel.jackson Jan 19 '11 at 9:25
I ended up writing it like this: $$|a_{n}-L|\leq |a_{n}-f_{n}(x)|+|f_{n}(x)-f(x)|+|f(x)-L|$$ and using similar arguments from before, got the desired result. Did you have something simpler in mind? – daniel.jackson Jan 21 '11 at 9:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.