# Fourier coefficient problem

1. May 15, 2005

### Zaare

I need help with the last part of the following problem:

Let $$f(x)$$ be a $$2\pi$$-periodic and Riemann integrable on $$[-\pi,\pi]$$.

(a) Assuming $$f(x)$$ satisfies the Hölder condition of order $$\alpha$$
$$\left| {f\left( {x + h} \right) - f\left( x \right)} \right| \le C\left| h \right|^\alpha ,$$​
for some $$0 < \alpha \le 1$$, some $$C > 0$$ and all $$x, h$$.
Show that
$${\hat f\left( n \right)} \le O ( \frac {1} { \left| n \right| ^ \alpha} )$$​
where $$\hat f\left( n \right)$$ is the Fourier koefficient.

(b) Proove that the above result cannot be improved by showing that the function
$$g\left( x \right) = \sum\limits_{k = 0}^\infty {2^{ - k\alpha } e^{i2^k x} }$$​
also with $$0 < \alpha \le 1,$$ satisfies
$$\left| {g\left( {x + h} \right) - g\left( x \right)} \right| \le C\left| h \right|^\alpha .$$​

I have done (a). And I was able to show that the sum satisfies the condition. What I don't understand is:
how can I use it to show that the result in (a) cannot be improved?
I thought that the sum can be iterpreted as the fourier series of $$g(x)$$, which would say that the Fourier coefficient of $$g(x)$$ is $$1/{n^\alpha}$$, where $$n=2^k$$. That is the resemblence I see between (a) and (b). But I don't know what to do with that.

Last edited: May 15, 2005
2. May 18, 2005

### Zaare

Ok, here's the way I reason:
In (a) I showed that the fourier coefficient has an upper limit. Then in (b) I showed that a certain function satisfying the desired condition has a Fourier coefficient which is of the same size as the upper limit in (a). Then obviously the upper limit cannot be improved.
Is it this obvious, or am I forgetting something?