# One-sided limits of a uniformly convergent function sequence and its limit function

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.

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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
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.
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