Regarding convolution of Fejér kernel with a Lipschitz function

Question

Let $\mathbb{T}$ be the unit circle in the complex plane. The Lipschitz class of functions on $\mathbb{T}$ is defined here. And the Fejér kernel is defined here. If $f\in Lip_{\alpha}(\mathbb{T}),\, 0<\alpha <1$, then how to show that $$\|K_n*f-f\|\leq C_{\alpha}\frac{\|f\|_{Lip_{\alpha}(\mathbb{T})}}{n^{\alpha}}$$ when $C_{\alpha}$ is an absolute constant depending on $\alpha$.

Answer 1

Hint: $$\begin{align}|K_n*f(x)-f(x)|&=\left|\frac1{2\pi}\int_{-\pi}^\pi K_n(t)(f(x-t)-f(x)\,dt\right| \\&\le\frac1{2\pi}\int_{-\pi}^\pi K_n(t)|f(x-t)-f(x)|\,dt.\end{align}$$

So now all you need is a suitable upper bound for $\frac1{2\pi}\int_{-\pi}^\pi K_n(t)|t|^\alpha=\frac1\pi\int_0^\pi K_n(t) t^\alpha$. Don't try to work out that integral exactly. Instead note that $\sin(t/2)\ge ct$ for $0\le t\le \pi$, so $$\begin{align}\int_0^\pi K_n(t)t^\alpha &\le \frac cn\int_0^\pi\frac{\sin^2(nt/2)}{t^2}t^\alpha \\&=\frac c{n^\alpha}\int_0^{n\pi}\frac{\sin^2(t/2)}{t^2}t^\alpha \\&\le \frac c{n^\alpha}\int_0^{\infty}\frac{\sin^2(t/2)}{t^2}t^\alpha;\end{align}$$note that the last integral is finite since $\alpha<1$.