Fourier Analysis an Intrduction Chapter 3 15 c

Question

Let $f$ be $2\pi$-periodic and Rienmann integrable on $[-\pi,\pi]$. $\hat f(n)$ is the fourier cofficient.

a) Show that $$\hat f(n)=-\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+\pi/n)e^{-inx}dx$$ hence $$\hat f(n)=\frac{1}{4\pi}\int_{-\pi}^{\pi}[f(x)-f(x+\pi/n)]e^{-inx}dx.$$

(b) Now assume that $f$ satisfies a Holder condition of order $\alpha$, namely $$|f(x+h)-f(x)|\le C|h|^{|\alpha|}$$ for some $0<\alpha\le 1$, some $C>0$, and all $x,h$. Use part a) to show $$\hat f(n)=O(\frac{1}{|n|^{\alpha}})$$

(c) Prove that the above result cannot be improved by showing that the function $$f(x)=\sum_{k=0}^{\infty}2^{-k\alpha}e^{i2^{k}x},$$ where $0<\alpha<1,$ satisfies $$|f(x+h)-f(x)|\le C|h|^{\alpha},$$ and $\hat f(N)=1/N^{\alpha}$ whenever $N=2^{k}.$

[Hint: For (c), break up the sum as follows $f(x+h)-f(x)=\sum_{2^{k}\le1/|h|}+\sum_{2^{k}>1/|h|}.$To estimate the first sum use the fact that $|1-e^{i\theta}|<|\theta|$ whenever $\theta$ is small. To estimate the second sum, use the obvious inequality $|e^{ix}-e^{iy}|\le 2$]

I have proved a) and b), but when I got to c), I was stuck with the proving. There is what I thought: $$|f(x+h)-f(x)|=|\sum_{k=0}^{\infty}2^{-k\alpha}e^{i2^{k}(x+h)}-\sum_{k=0}^{\infty}2^{-k\alpha}e^{i2^{k}x}|=|\sum_{k=0}^{\infty}2^{-k\alpha}e^{i2^{k}x}(e^{i2^{k}h}-1)|$$

split this into $f(x+h)-f(x)=\sum_{2^{k}\le1/|h|}+\sum_{2^{k}>1/|h|}$ and use the hint: $$|\sum_{k=0}^{\infty}2^{-k\alpha}e^{i2^{k}x}(e^{i2^{k}h}-1)|\le |\sum_{2^{k}\le1/|h|}2^{-k\alpha}e^{i2^{k}x}2^{k}h| + |\sum_{2^{k}>1/|h|}2^{-k\alpha}e^{i2^{k}x}2| $$

$$\le \sum_{2^{k}\le1/|h|}2^{-k\alpha} + 2 \sum_{2^{k}>1/|h|}2^{-k\alpha}$$ $$\le \sum_{2^{k}\le1/|h|}|h|^{\alpha}+2 \sum_{2^{k}>1/|h|}|h|^{\alpha}$$ But I don't know what's the next step, and I don't know how does this prove shows that the result in b) cannot be improved.

Answer 1

You are correct up to the step $$\sum_{2^{k}\le1/|h|}2^{-k\alpha} + 2 \sum_{2^{k}>1/|h|}2^{-k\alpha}$$To keep it going, note that the first term ($\sum_{2^{k}\le1/|h|}2^{-k\alpha}$) is bounded (since it is a summation over some terms bounded in number) and can be considered as a constant. Now for the second term, we exploit the infinite geometric sum formula as:$$2\cdot\sum_{2^{k}>1/|h|}2^{-k\alpha}{=2\cdot {t_0\over 1-q}\\={2t_0\over 1-2^{-\alpha_0}}}$$but what is $t_0$? We can't answer this question precisely, but we can bound it. Note that since the sum is calculated over $2^k>{1\over |h|}$ then we have $$t_0={2^{-k_0\alpha}}<{h^\alpha}$$so finally by plugging this result in the second term we can finally write$$\sum_{2^{k}\le1/|h|}2^{-k\alpha} + 2 \sum_{2^{k}>1/|h|}2^{-k\alpha}\le K+{2\over 1-2^{-\alpha}}\cdot h^\alpha$$hence $\hat f(n)$ can be calculated easily.