At Stein's book of Fourier analysis (Chapter 3, page 91, exercise 15) I was trying to solve the following problem I have to prove that the result
$$\widehat{f}(n)=O\left(\frac{1}{|n|^{\alpha}}\right)$$
Cannot be improved by showing that the function:
$$f(x)=\sum_{k=0}^{\infty}2^ {−kα} e^{i2^{k}x}$$
with $0<α<1$, satisfies
$$|f (x + h) − f (x)| ≤ C|h|^{α}$$
and $\widehat{f}(N ) = 1/N^{α}$ whenever $N = 2^k$
But I don't know what I am supposed to do here?, there is a hint but it doesn't clarify the question, Can someone help me with this question please?
Best Answer
This is a Weierstrass-type function. One can prove the continuity by splitting the sum into low frequencies and high frequencies; the cutoff will depend on the scale $h$. Namely, $k_0 = -\log_2 h$.
High frequencies. If $k\ge k_0$, then $2^{-k}\le h$, the terms are small enough so that we can simply bound them by the triangle inequality: they contribute $O(h^\alpha)$.
Low frequencies. If $k<k_0$, then the derivative of the $k$th term, namely $i 2^ {k(1-\alpha)} e^{i2^{k}x}$, is bounded by $2^ {k(1-\alpha)} $. Summing this over $k\le k_0$ shows the derivative of the sum of all low-frequency terms is $O(2^{k_0(1-\alpha)})$ which is $O(h^{\alpha-1})$. Hence, the difference of values of this sum on the scale of $h$ is $O(h^\alpha)$.