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

## 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 :)

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The thing you always need to do first is to determine the pointswise limit. To what function does $$\cos^{2n}(x)$$ converge pointswise? Is this limit continuous?

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.

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...

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.

Why do you want to do things with the derivative so much. It'll just make things complicated

All of the $$\cos_{2n}(x)$$ are continuous, yet the limit is not continuous. Hence the sequence does not converge uniformly...

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.

Oh I see. I understand now...

What you said in post 3 is the correct way of showing this thing I think...

Ok cool, thanks for that :) sorry about the confusion