Can Fourier Coefficient Bounds Under Hölder Conditions Be Improved?

[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.

 

[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?

 

[thepatientmental]

In order to show that the result in (a) cannot be improved, we need to show that the function g(x) given in (b) does not satisfy a stronger condition than the Hölder condition of order α. This means that the Fourier coefficients of g(x) cannot decay faster than O(1/n^α), as shown in (a).

As you correctly noted, g(x) can be interpreted as the Fourier series of itself, with the Fourier coefficients being 1/n^α where n = 2^k. This means that the sum given in (b) is actually the Fourier series of g(x), and we can use this to compare the Fourier coefficients of g(x) with those of f(x).

Since f(x) satisfies the Hölder condition of order α, we know that its Fourier coefficients decay at least as fast as O(1/n^α). However, for g(x), we can see that the Fourier coefficients decay exactly at the rate of 1/n^α. This means that the Hölder condition of order α is the strongest possible condition that g(x) can satisfy, and it cannot be improved upon.

In other words, g(x) is an example of a function that satisfies the Hölder condition of order α, but does not satisfy any stronger condition. Therefore, the result in (a) cannot be improved, as g(x) provides a counterexample to any potential stronger condition.