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Everywhere continuous, nowhere differentiable function
By Yatir Halevi on Tuesday, October 29, 2002 - 04:08 pm:

Why is the Weierstrass Function http://mathworld.wolfram.com/WeierstrassFunction.html not differentiable?

Yatir

By Andre Rzym on Tuesday, November 05, 2002 - 09:07 am:

I found an article which made reference to a paper by Hardy (around 1912) which shows this function to be non-differentiable for ab>1. Does anyone have access to his collected papers? I have one volume of them but it's not in there.

Andre

By Ian Short on Tuesday, November 05, 2002 - 01:26 pm:

I think the result is not true for ab £ 1 and Hardy or someone established this too.

(1) Yitzhak Katznelson proves the result for large a and b on page 105 of "Introduction to Harmonic Analysis".

(2) Tom Korner gives a simple proof with bn and an replaced by factorials in his book "Fourier Analysis".

The key point in both cases being that the theorem is easier to prove when the a and b terms are large and small respectively. Weierstrass originally proved the result with large a and small b and Hardy + others found the best bounds as Andre points out (I think).

The proof is technical, Upwards more than Onwards, but the idea is more or less as follows (at least for Tom's proof).

The first term in the series is a periodic graph. Add on the next term and this is also periodic, but with higher frequency and smaller amplitude. It makes the original periodic graph "quiver". As you add more cos terms on, the graph quivers more and more and the gradients get higher. That is, the maximum of the derivative of each partial sum increases without bound rendering the limit function nowhere differentiable.

I could well supply details (taken from the books) or some pictures if anyone is still with this. Tom Korner supplied me with all the information, although I might have made some slips above regarding the history of the function.

Ian

By Ian Short on Wednesday, November 06, 2002 - 03:12 pm:

The simplest example I have now encountered is the series,

a(x)=Sn=0¥ (1/n!)sin[(n!)2x]
=aN(x)+bN(x)+cN(x)

where aN, bN and cN represent the (N-1)th partial sum, the Nth term and the remaining sum from n=N+1 to infinity, respectively. I.e. we have split the series in three. The (continuous) partial sums of this series converge uniformly to the continuous function a.

Fix a real number x and we construct a sequence x1,x2,... convergent to x which lies within a bN period of x so that |aN(x)-aN(xN)| is small (as xN is close to x) and |cN(x)-cN(xN)| is small, relative to |bN(x)-bN(xN)|. I'll give most details, but leave some algebraic steps.

%%%%% It's possible to choose p/(N!)2 < |x-xN| £ 3p/(N!)2 with |bN(x)-bN(xN)| ³ 1/N!.

%%%%% Use |sinx-siny| £ |x-y| to show |aN(x)-aN(xN)| £ 6p/N(N!).

%%%%% |cN(x)-cN(xN)| £4/(N+1)!.

%%%%% Hence, |a(x)-a(xN)|
³ |bN(x)-bN(xN)|-|aN(x)-a N(xN)|-|cN(x)-cN(xN)|
³ 1/2N! for large enough N.

Therefore |a(x)-a(xN)|/|x-xN| ³ N!/6p.

That seemed a bit too easy!! The choice of xN seems to do most of the work.

Ian