This problem is from Stein & Shakarchi's Fourier Analysis, Exercise 16(b) of Chapter 3.
Let $f$ be a $2\pi$-periodic function which satisfies a Lipschitz condition with constant $K$; that is, $$|f(x)-f(y)|\leq K|x-y|$$ for all $x,y.$
Let $p$ be a positive integer and $g_h(x)=f(x+h)-f(x-h)$ for $h>0$. By choosing $h=\pi/2^{p+1}$, show that $$\sum_{2^{p-1}<|n|\leq 2^p}|\hat{f}(n)|^2\leq \frac{K^2\pi^2}{2^{2p+1}}.$$
My attempt: I already proved from the previous sub-problem 16(a) that $$\frac{1}{2\pi}\int_0^{2\pi}|g_h(x)|^2dx=\sum_{n\in\mathbb{Z}}4|\sin nh|^2|\hat{f}(n)|^2\leq 4K^2h^2$$ using Parseval's identity with $|\hat{g_h}(n)|=2|\sin nh||\hat{f}(n)|.$
To estimate $|\hat{f}(n)|^2$ for $2^{p-1}<|n|\leq 2^p$, I proceeded as follows: $$ |\hat{f}(n)|^2=\frac{|\hat{g_h}(n)|^2}{4|\sin nh|^2}\leq \frac{|\hat{g_h}(n)|^2}{2}\leq \frac{1}{2}\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x+h)-f(x-h)|dx\right)^2\\\leq \frac{1}{4\pi}\int_{-\pi}^{\pi}|f(x+h)-f(x-h)|^2dx\leq \frac{1}{4\pi}\int_{-\pi}^{\pi}4K^2h^2dx=2K^2h^2\\=\frac{K^2\pi^2}{2^{2p+1}} $$
However, I want to get the upper bound $\frac{K^2\pi^2}{2^{2p+1}}$ for the sum of the squares of Fourier coefficient, not the individual term. Does anyone have ideas to refine this bound for the desired upper bound?
Thanks in advance!
Note that $h=\pi/2^{p+1}$ implies that $|\sin nh|^2\geq 1/2$ for $2^{p-1}<|n|\leq 2^p$. Therefore, by Parseval's identity, $$\frac{1}{2}\sum_{2^{p-1}<|n|\leq 2^p}|\hat{f}(n)|^2\leq \sum_{2^{p-1}<|n|\leq 2^p}|\sin nh|^2|\hat{f}(n)|^2\leq \sum_{n\in \mathbb{Z}}|\sin nh|^2|\hat{f}(n)|^2\\=\frac{1}{8\pi}\int_{-\pi}^{\pi}|g_h(x)|^2dx\leq \frac{1}{8\pi}\int_{-\pi}^{\pi}K^2(2h)^2dx={K^2h^2}=\frac{K^2\pi^2}{2^{2p+2}}$$ and completes the proof.