- Let $f$ be a $2\pi$-periodic function which satisfies a Lipschitz condition with constant $K$ that is,
$$|f(x)-f(y)| \le K|x-y|,\ for\ all\ x,y.$$
This is simply the Holder condition with $\alpha =1$: $$|f(x+h)-f(x)| \le C|h|^{\alpha},\ C>0, 0<\alpha \le 1$$ so by the previous exercise, we see that $\hat f(n)=O(1/|n|).$ Since the harmonic series $\sum 1/n$ diverges, we cannot say anything(yet) about the absolute convergence of the Fourier series of $f.$ The outline below actually proves that the Fourier series of $f$ converges absolutely and uniformly.
(a) For every positive $h$ we define $g_{h}(x)=f(x+h)-f(x-h).$ Prove that $$\frac{1}{2\pi}\int_{0}^{2\pi}|g_{h}(x)|^{2}dx=\sum_{n=-\infty}^{\infty}4|sin(nh)|^{2}|\hat f(n)|^{2},$$
and show that
$$\sum _{n=-\infty}^{\infty}|sin(nh)|^{2}|\hat f(n)|^{2} \le K^{2}h^{2}.$$ (b) Let $p$ be a positive integer. By choosing $h=\pi /2^{p+1},$ show that $$\sum_{2^{p-1}<|n|\le 2^{p}}|\hat f(n)|^{2} \le \frac{K^{2}\pi^{2}}{2^{2p+1}}.$$ (c) Estimate $\sum_{2^{p-1}<|n|\le 2^p}|\hat f(n)|,$ and conclude that the Fourier series of $f$ converges absolutely, hence uniformly. [Hint: Use the Cauchy-Schwarz inequaity to estimate the sum.]
I solved (a) and (b) and proved the convergence of by the following process: $$\sum_{2^{p-1}<|n|\le 2^p}|\hat f(n)| \le \sum_{2^{p-1}<|n| \le 2^p}|\hat f(n)|^{2}$$ $$\sum_{n=-\infty}^{\infty}|\hat f(n)| \le \sum_{p=1}^{\infty}\frac{k^{2}\pi^{2}}{2^{2p+1}}=\frac{k^{2}\pi^{2}}{6}$$ but I don't know how to use the Cauchy-Schwarz inequality to estimate the sum.
I can't see how you justify that $\sum_{2^{p-1}\leq|n|\leq 2^p}|\hat f(n)|\leq\sum_{2^{p-1}\leq|n|\leq 2^p}|\hat f(n)|^2$. If the series will converge, it has to be false, because eventually $|\hat f(n)|<1$.
I think what they want you to do is to use (b) via \begin{align} \sum_{2^{p-1}\leq|n|\leq 2^p}|\hat f(n)|&\leq\left(\sum_{2^{p-1}\leq|n|\leq 2^p}|\hat f(n)|^2\right)^{1/2} \left(\sum_{2^{p-1}\leq|n|\leq 2^p}1^2\right)^{1/2}\\ &=\left(\sum_{2^{p-1}\leq|n|\leq 2^p}|\hat f(n)|^2\right)^{1/2} (2(2^p-2^{p-1}))^{1/2}\\ &=2^{p/2}\left(\sum_{2^{p-1}\leq|n|\leq 2^p}|\hat f(n)|^2\right)^{1/2} \\ &\leq 2^{p/2}\,\frac{K^2\pi^2}{2^{2p+1}}\\ &=\frac{K^2\pi^2}{2^{3p/2+1}}. \end{align} Now you can show that the tails of the series are arbitrarily small.