Theorem - Uniform convergence

[estro]

1. The problem statement, all variables and given/known data

 

Suppose:

 

[math]f(x)\ and\ f'(x)\ are\ continuous\ for\ all\ x \in R [/math]

 

[math]For\ all\ x \in R\ and\ for\ all\ n \in N[/math]

[math]f_n(x)=n[f(x+\frac{1}{n})-f(x)][/math]

 

[math]Prove\ that\ when\ a,b\ are\ arbitrary,\ f_n(x)\ is\ uniform\ convergent\ in\ [a,b][/math]

 

 

3. The attempt at a solution

 

 

[math]\lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty} n[f(x+\frac{1}{n})-f(x)] = \lim_{t\rightarrow0} \frac {f(x+t)-f(x)}{t}=f'(x)[/math]

 

[math]\max_{[a,b]}|n[ f(x+\frac{1}{n}) -f(x)]-f'(x)|=|n[ f(x_0+\frac{1}{n}) -f(x_0)]-f'(x_0)|=[/math][math]|\frac {f(x_0+t)-f(x_0)}{t}-f'(x_0)|\rightarrow0[/math]

 

I fear that I miss something terribly important.

*I left out all the little technical details to make things shorter.

[Dave]

Good problem! I'd originally thought you weren't along the right lines, but now I've studied things a little better (and jogged my memory regarding supremums ) I can see the logic. As far as I can tell, it simply needs tidying up. Here's my version.

 

Since [imath]f_n[/imath] and [imath]f[/imath] are continuous, it follows that [imath]f_n-f[/imath] is continuous on [imath][a,b][/imath], and hence there exists [imath]x_0[/imath] such that

 

[math]\sup_{x\in [a,b]} |f_n(x)-f(x)| = |f_n(x_0)-f(x_0)|[/math].

 

Then,

 

[math]\lim_{n\to\infty} \sup_{x\in [a,b]} |f_n(x)-f(x)| = \lim_{t\to 0} \left| \frac{f(x_0+t) - f(x_0)}{t} - f'(x_0)\right| = 0[/math],

 

so [imath]f_n[/imath] converges uniformly to [imath]f[/imath].

Edited June 16, 2010 by dave

[estro]

[math]\mbox{f(x) and f'(x) are continuous for all x in R. (*1)} [/math]

 

[math]\mbox {Let a,b in R\ so\ that\ without\ the\ loss\ of\ generality } a<b.[/math]

 

[math]\mbox {Let x in [a,b]. (*2)} [/math]

 

[math]\lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty} n[f(x+\frac {1} {n})-f(x)]=[/math][math][t_n=1/n]=\lim_{t\rightarrow0} \frac {f(x+t)-f(x)} {t}=f'(x). \ (*3)[/math]

 

[math]\mbox{(*1) and (*2) and (*3) } \Rightarrow\ g(x)=|f_n(x)-f(x)| \mbox{ is continuous for all x in R.\ (*4)}[/math]

 

[math]\mbox{(*2) and (*4)} \Rightarrow\ \mbox {Weierstrass Theorem } \Rightarrow\[/math][math] \exists\ \max_{[a,b]} g(x)=g© \mbox { c in [a,b].\ (*5)} [/math]

 

[math]\mbox {(*5) and (*1) } \Rightarrow\ \forall\ x\in[a,b]\ \Rightarrow\ \sup_{[a,b]}|f_n(x)-f(x)|=|\frac {f(c+t)-f©} {t} -f'©|\rightarrow0[/math]

 

[math]\Rightarrow \mbox { Basic lemma for uniform convergence }[/math][math] \Rightarrow\ f_n(x) \mbox{ is uniform convergent in [a,b], where a,b are arbitrary.} [/math]

 

QED

 

[Edit] "Basic lemma for uniform convergence" is sounds little funny but I'm sure you'll understand what I meant.

[Edit] I'm afraid of having the wrong intuition about uniform convergence, I hope you will be be able to say if I understand the idea from my proof.

[Edit] If only I could prove g(x) is monotonic [such thing is possible?] then my proof would be a lot shorter, using Dini Theorem.

Edited June 17, 2010 by estro