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).
Uniform convergence and differentiation
2 Answers
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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0@ThomasAndrews: I hope it's explicit now. What would you say? – 2011-10-19
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. Let $n_0$ be the least $n$ such that $v+h_n, 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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0Yeah, 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. – 2011-10-19