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Let $f$ be a continuous function on $[-1,2]$. Given $0\le x\le 1$ and $n\ge 1$, define a sequence of functions: $$f_n(x)=\frac{n}{2}\int\limits_{x-\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}\,.$$ Show that each $f_n$ is continuous on $[0,1]$ and that $(f_n)$ converges uniformly to $f$ on $[0,1]$.

My work on continuity:

Let $\epsilon>0$. Let $x_n\to x$ in $[0,1]$. Let $\delta$ be the continuity criterion for $f$ (not sure what else to call it); then there exists $N$ such that $|x_n-x|<\delta$ for $n\ge N$. For such $n$, we thus have $|f(x_n)-f(x)|<\epsilon$. So for $t\in (x-\delta, x+\delta)$, we have $|f(t)|<|f(x)|+\epsilon$.

Then

$$|f_n(x_n)-f_n(x)|=\left|\frac{n}{2}\int\limits_{x_n-\frac{1}{n}}^{x_n+\frac{1}{n}}{f(t)\,dt}-\frac{n}{2}\int\limits_{x-\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}\right|$$

Now, I don't want to typeset all the madness I have scribbled down, but this difference of integrals becomes (if we assume $x_n<x$ and forget the $n/2$):

$$\int\limits_{x_n-\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}-\int\limits_{x_n+\frac{1}{n}}^{x+\frac{1}{n}}{f(t)\,dt}$$

Then, using the fact that $f(t)<f(x)+\epsilon$ over each of these intervals, everything whittles down to being $$\le n\cdot |f(x)+\epsilon|\cdot (x-x_n)\,.$$

But... uhhh... ? Any help you can give is appreciated

EDIT: What I want to end up with is $|f_n(x_n)-f_n(x)|<\epsilon$, but I don't see how to get there from what I have. Or can I get that from what I have? Am I on the right path?

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What is your question? –  Rasmus Apr 18 '11 at 17:27
    
@Rasmus: How do I get $|f_n(x_n)-f_n(x)|<\epsilon$ from what I have? Or can I, even? I have added this to the end of my post. –  Bey Apr 18 '11 at 17:52
1  
you're quite close in fact. Beware that you have two different meanings of $n$ in your argument after "Then"! –  t.b. Apr 18 '11 at 17:55
    
@Theo: Whoops! My mistake. An earlier homework problem involved both a sequence of functions and a sequence of points, and in that problem the indices were supposed to line up. Thanks for pointing that out. –  Bey Apr 18 '11 at 18:16
2  
Well, this happens to all of us. Eric has elaborated on that in his answer below. But you'll see that you were not that far from a solution (at least for continuity of $f_n$). –  t.b. Apr 18 '11 at 18:18

1 Answer 1

up vote 3 down vote accepted

Ok, this is very confusing for me to read. One big problem is that you use the index $n$ to refer to two different things simultaneously. You use it both for the sequence $f_n$ and for the sequence $x_n$. With this small change, everything you wrote is correct, and what follows is not much different:

Now, the idea is we fix $k$ and look at $f_k$ and we want to show that $f_k$ is continuous. As you wrote, $$|f_k(x_n)-f_k(x)|=\left|\frac{k}{2}\int\limits_{x_n-\frac{1}{k}}^{x_n+\frac{1}{k}}{f(t)\,dt}-\frac{k}{2}\int\limits_{x-\frac{1}{k}}^{x+\frac{1}{k}}{f(t)\,dt}\right|.$$ The key idea is that if $k$ is fixed, we can choose $\delta $ really really small so that these integrals almost line up entirely and cancel out. So, assume $x_n<x$ and $|x-x_n|<\delta$ and say $\delta <\frac{1}{k}$. Then

$$\left|\frac{k}{2}\int\limits_{x_n-\frac{1}{k}}^{x_n+\frac{1}{k}}{f(t)\,dt}-\frac{k}{2}\int\limits_{x-\frac{1}{k}}^{x+\frac{1}{k}}{f(t)\,dt}\right|= \left|\frac{k}{2}\int\limits_{x_n-\frac{1}{k}}^{x-\frac{1}{k}}{f(t)\,dt}+\frac{k}{2}\int\limits_{x_n+\frac{1}{k}}^{x+\frac{1}{k}}{f(t)\,dt}\right|.$$ But these two integrals are over an interval of length $\delta$. Since $f$ must be bounded on $[-1,2]$ we can choose $M$ such that $M>|f(x)|$ for all $x$. Then the above term is strictly less than $$\frac{2\delta kM}{2}=kM\delta. $$ Since both $k$ and $M$ are fixed constants, we can choose $\delta$ small enough so that this is less than a given $\epsilon.$

Hope that helps,

Short Answer: Not sure why, but I rewrote the whole proof above. Here is the short answer to your question:

You had $n |x-x_n| |f(x)+\epsilon|$ as an upper bound. Replacing the $n$ with a $k$, to distinguish the two sequences, using an upper bound for $f(x)$ on the whole interval and using the fact that $|x-x_n|<\delta$ this upper bound becomes $$k\delta M.$$ Then as $k,M$ are fixed, choose $\delta =\epsilon/kM$ and it is finished.

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Ah, thanks! It hadn't occurred to use the compactness of $f$'s domain to bound the silly thing. Also, as I told Theo above, an earlier homework problem involved both a sequence of functions and a sequence of points with the indices matching, and obviously I forgot to clear my head. =) –  Bey Apr 18 '11 at 18:18

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