Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g'(x)$.

share|improve this question

2 Answers 2

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 $I$.

share|improve this answer
    
$n$ 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 $n$.
share|improve this answer
    
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

Your Answer

 
discard

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.