(Dis)proof of a sequence of functions being uniformly convergent

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Homework Statement



Let g_n : R -> R be given by gn (x) := cos2n (x), does gn' converge uniformly?


Homework Equations



The derivative is as follows; -2nsin(x)cos2n-1, which I have found converges pointwise to the 0 function.

Formal definition of Uniform Convergence;

For all e>0, there exits N(e) in the natural numbers, such that for all n >= N, for all x in R,
imples Mod( fn(x) - f(x)) < e
where f is the limit of the sequence of functions fn


The Attempt at a Solution



I have been struggling at seeing how to prove functions are uniformly convergent in general, but have been told that this does not converge uniformly, however I really cannot see how this is true, as

Supremum norm(gn' - 0) ->0 as n-> Infinity,

which is an equivalent definition of uniform convergence, because the term cos2n-1 is dominant over 2n.

Thanks for any help :)
 

Answers and Replies

  • #2
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The thing you always need to do first is to determine the pointswise limit. To what function does [tex]\cos^{2n}(x)[/tex] converge pointswise? Is this limit continuous?
 
  • #3
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Sorry, I omitted my work there, as did not think it was as relevant, I found the pointwise limit, say g, to be defined as;

g(x) := {0 for x =/= k*Pi
{1 for x = k*Pi

For some k in the integers

This shows the limit function is not continuous, I had thought about using the theorem which says when gn is a sequence of C1 functions, that converge pointwise to g. Suppose gn' does converge uniformally, then it would have to converge to f, then g is C1 and f = g'.

Thus this would arrive at a contradiction as g is not C1, but I am not sure if you can twist the theorem in this way.
 
  • #4
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Yes, I suppose you could use that. But you're making things awfully complicated...

You probably showed somewhere that if a sequence of continuous functions converges uniformly, then it's limit is continuous to. I suggest you use this...
 
  • #5
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How could that be applicable to the derivative of the function though, I do not remember anything saying that if the sequence is not uniformly convergence, then the derivative must not be uniformly convergent.

The only thing I can think of is looking at the theorem about a sequence of regulated functions converging uniformly to f, then the integral of the sequence must converge to the integral of f.
 
  • #6
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Why do you want to do things with the derivative so much. It'll just make things complicated :confused:

All of the [tex]\cos_{2n}(x)[/tex] are continuous, yet the limit is not continuous. Hence the sequence does not converge uniformly...
 
  • #7
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Sorry I thought my function was clearer :P the question is asking whether the derivative converges, not the main function, hence why I was putting the ' next to some g, but looking back at it it is quite unclear hehe.
 
  • #8
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Oh I see. I understand now...

What you said in post 3 is the correct way of showing this thing I think...
 
  • #9
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Ok cool, thanks for that :) sorry about the confusion
 

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