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Consider a sequence of functions $f_n$ where $f_n : \mathbb{R} \to \mathbb{R}$ and $f_n$ are all differentiable with derivatives $f^\prime_n$. The sequence $f_n$ and the sequence $f^\prime_n$ both converge uniformly to functions $f$ and $g$ respectively. According to the definition given in this wiki page on uniform convergence in section 'To Differentiability"

If $ f_n $ converges uniformly to $ f $, and if all the $ f_n $ are differentiable, and if the derivatives $f^\prime_n$ converge uniformly to $g$, then $ f $ is differentiable and its derivative is $g$.

Q.1) Can a similar definition be used for higher order derivatives ?

This is a different definition of derivative which is not same as the usual definition $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$.

Q.2) In case of both not being in agreement with each other which one should I use for further analysis on $f$ for example in theorems involving the derivative of $f$ ?

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That's not a definition. It's a theorem. – Chris Eagle Jan 21 '11 at 8:25
@Chris Eagle : what if $f$ is not differentiable in conventional sense. – Rajesh Dachiraju Jan 21 '11 at 8:26
You seem to be confusing a theorem for a definition. And in Q.1, what do you mean by higher order derivatives? Usually that just means taking derivatives of derivatives, i.e., $f''=(f')'$ is the second derivative, $f'''=(f'')'$ is the third derivative, etc. If you have uniform convergence for all of these, then a similar theorem will hold, where the higher order derivatives of the limit will be the limit of the higher order derivatives. But I don't know what you're asking. You say "this is a different definition", but you haven't said what "this" is; there is no other definition. – Jonas Meyer Jan 21 '11 at 8:26
@Jonas Meyer : there are two ways of deriving derivative of $f$ by using the definition $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ and getting directly or by using the uniform convergence of $f^\prime_n$ to $g$. What if both are not equal ? – Rajesh Dachiraju Jan 21 '11 at 8:30
$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ is the definition. Part of the conclusion of the theorem is that $f'$ exists, and that $f'=g$. That is, $f'$ is not defined to be $g$; it already has a definition, when it exists, and the theorem asserts that under those hypothesis is does exist, and it also equals $g$. – Jonas Meyer Jan 21 '11 at 8:34
up vote 2 down vote accepted

If $f_n$ converges uniformly to $f$, and if all the $f_n$ are differentiable, and if the derivatives $f'_n$ converge uniformly to $g$, then $f$ is differentiable and its derivative is $g$.

So if you use the theorem and the derivates $f'_n$ are also differentiable you have again that the $f'_n$ converge uniformly to $f'$ and you can recursively use the theorem from above (you will need that $f''_n$ converges uniformly to some $g'$).

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The theorem you mention certainly applies to higher order derivatives, exactly as written by user3123 above.

For your other question, it seems to me that you are worried about the situation where $f$ may not be differentiable, but there exists a sequence $f_n$ which uniformly converges to $f$, with each $f_n$ differentiable and the derivatives $f_n'$ converging uniformly to some function $g$.

Do you have an example of such a situation?

To rephrase what Jonas mentioned already in his comments, this is not likely to happen. When the sequence $f_n$ uniformly converges to $f$, and the derivatives $f_n'$ uniformly converge to $g$, the following holds:

$\lim_{h\searrow 0}\frac{f(x+h)-f(x)}{h} = g(x).$

This is the derivative of $f$. It is not an alternative definition, it really is a theorem. I'm sorry! Perhaps you can give some more details on the exact situation you have in mind.

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I was far from anticipating such a situation apriori to being caught in it. Thats why my optimism made me to misjudge a theorem to be a definition. I've cross checked it, the convergence is uniform for the functions and its derivatives but the limit function is clearly not differentiable. Also there seem to be no dependency between limit function $f$ and limit of derivatives $g$. – Rajesh Dachiraju Jan 21 '11 at 13:05
I sincerely do not doubdt the theorem, even though i know very little about it. – Rajesh Dachiraju Jan 21 '11 at 13:10
May I ask, what is the sequence of functions exactly? Perhaps this will help clarify the point at hand. – Glen Wheeler Jan 21 '11 at 14:19
@Rajesh: you have already been asked for this sequence you are working with before. Since you should now be fully aware that the situation you describe is impossible, you have made a mistake. So it would be easier on all of us if you posted the sequence and let us find the mistake. My guess is that the derivatives don't converge uniformly; this is the easiest mistake to make. – Qiaochu Yuan Jan 21 '11 at 14:23
@Qiachu Yuan : I found my mistake...thank you...I guess converse need'nt be is not a necessary condition for $f$ to be differentiable. – Rajesh Dachiraju Jan 22 '11 at 5:18

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