# Uniform convergence and differentiation

How does one show that given that $g\in C^1(a,b)$, the sequence of functions $$g_n=n\left(g\left(x+{1\over n}\right)-g(x)\right)$$ converges uniformly on all closed intervals in $(a,b)$? I assume the limit function is $g'(x)$.

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Ok, so one of the proof is the following. Let $I$ be a closed subinterval of $(a,b)$ so $g'\in C(I)$ and hence uniformly continuous on $I$. Let us put $n_0$ such that $I' = I\cup\{x+1/n:x\in I\}\subset (a,b)$. By Lagrange Mean Value Theorem for $n>n_0$ we have: $$g_n(x) = g'(\xi_n(x))$$ where $\xi_{n}(x)\in[x,x+1/n]$.

Since $g'$ is uniformly continuous on a closed subinterval $I'\subset(a,b)$ it means that for any $\epsilon>0$ there is $\delta(\epsilon)>0$ such that if $|x-y|<\delta(\epsilon)$ where $x,y\in I'$ it holds that $|g'(x)-g'(y)|<\epsilon$.

That means, that for any $\epsilon>0$ there is $N>\max\left\{n_0,\frac{1}{\delta(\epsilon)}+1\right\}$ such that for any $n>N$ it holds that $$\sup\limits_{x\in I}|g_n(x) - g'(x)|\leq\sup\limits_{x\in I'}|g'(\xi_n(x)) - g'(x)|<\epsilon$$ which proves the uniform convergence $g_n\to g'$ on $I$.

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$n$ small enough? Don't you mean large enough? – Jonas Teuwen Oct 19 '11 at 14:54
@JonasTeuwen: sure, thanks ) – Ilya Oct 19 '11 at 14:56
You have to be a little careful to expand your closed interval, since $x$ being in your closed interval doesn't mean that $x+\frac{1}{n}$ is in your interval. So you need the fact that $g'$ is uniformly continuous on some expanded closed interval still contained in $(a,b)$. – Thomas Andrews Oct 19 '11 at 16:16
@ThomasAndrews: I assumed that $n$ is large enough meaning exactly that argument to apply MVT. Thank you for the comment, I maybe should mention it explicitly. – Ilya Oct 19 '11 at 16:21
But you have to specify the interval on which $g'$ is uniformly continuous for your argument to work. You can't just use that $g'$ is uniformly continuous on the question interval, it needs to be on some expanded interval. – Thomas Andrews Oct 19 '11 at 16:30

Starting fresh.

Letting $h_n=\frac{1}{n}$, then $g_n(x) = \frac{g(x+h_n) - g(x)}{h_n}$

For any $x$, by the mean value theorem, $g_n(x) = g'(y)$ for some $y\in [x,x+h_n]$.

Let $C=[u,v]$ be our closed interval. Note that $g_n$ is defined on $C$ only if $v+h_n<b$. Let $n_0$ be the least $n$ such that $v+h_n<b$, and let $C'=[u,v+h_{n_0}]$.

Since $C'$ is compact, and $g'$ is continuous, $g'$ must be uniformly continuous on $C'$. Therefore, given $\epsilon>0$ there is a $\delta>0$ such that if $|x-y|<\delta$, $|g'(x)-g'(y)|<\epsilon$ for $x,y\in C'$.

Choose $N>\max(n_0,\frac{1}{\delta})$. Then for every $x\in C$, and any $n>N$, there is a $y$ in $[x,x+h_n]\subset C'$ such that $g_n(x)=g'(y)$. But then, since $x,y\in C'$:

$$|g_n(x)-g'(x)| = |g'(y)-g'(x)| < \epsilon$$

since $|y-x| < h_n < \frac{1}{N} < \delta$.

So the $g_n$ converge uniformly to $g'$.

So this essentially results from:

• The intermediate value theorem when derivatives are continuous
• The fact that a continuous function on a compact set is uniformly continuous on that set
• Currently, the proof uses that $h_n$ decreases and has limit 0, but you can prove it with general $h_n\rightarrow 0$, even allowing some $h_n=0$ if you define $g_n(x)=g'(x)$ for those $n$.
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Did you mean Dini's theorem: en.wikipedia.org/wiki/Dini's_theorem ? See then example (c) here: math.ubc.ca/~feldman/m321/dini.pdf – Ilya Oct 19 '11 at 15:04
@Gortaur Yeah, I might be misremembering my real analysis. – Thomas Andrews Oct 19 '11 at 15:16
Okay, fixed proof. Misremembered theorem about uniform continuity to be about uniform convergence. Duh. @Gortaur – Thomas Andrews Oct 19 '11 at 15:46
Nice, though seems to be similar to my answer ) – Ilya Oct 19 '11 at 15:48
Yeah, I got a wrong start, and continued working, only to see your answer afterwards. Kept mine because I had the part which expanded the interval. – Thomas Andrews Oct 19 '11 at 16:27