Show that $\sup_{x \in \mathbb{R}}|\sigma_nf(x) - f(x)| \leq C\frac{\log n}{n}$, for $f$ $2\pi$-periodic and Lipschitz.

Question

I'm learning about Fourier analysis and need help with the following 2 problems:

$(1)$ Show that $\forall t \in (0, \pi], K_n(t) \leq \min\{n +1, \frac{\pi^2}{(n + 1)t^2}\}$.

$(2)$ For the $2\pi$-periodic function $f,\exists K > 0$ such that $\forall x, y \in \mathbb{R}, |f(x) - f(y)| \leq K|x - y|$. Show that $\sup_{x \in \mathbb{R}}|\sigma_nf(x) - f(x)| \leq C\frac{\log n}{n}$, where $C =$ constant $> 0$.

Symbolism: $D_n(t)$ and $K_n(t)$ is the Dirichlet and Fejér kernel, respectively. $\sigma_nf(x)$ is the $n^{th}$ Cesàro mean of the Fourier series, that is $\sigma_nf(x) = \frac{S_0f(x) + \cdots + S_nf(x)}{n + 1}$, where $S_nf(x)$is the partial sum of the Fourier series.

My work:

$(1)$ By the definition of the Fejér kernel, for $n \in \mathbb{N}$ and $t \in \mathbb{R}$ we have

$$K_n(t) = \frac{1}{n + 1} \sum_{k = 0}^{n} D_k(t) = \frac{1}{n + 1} \Big(1+ \sum_{k = 1}^{n} D_k(t) \Big)$$

$$ = \frac{1}{n + 1} \Big(1 + \sum_{k = 1}^{n} \big(1 + 2\sum_{j = 1}^{k} \cos(jt) \big) \Big) \leq \frac{1}{n + 1} \Big(1 + \sum_{k = 1}^{n} \big(1 + 2\sum_{j = 1}^{k} 1 \big) \Big)$$

$$= \frac{1}{n + 1} \Big(1 + \sum_{k = 1}^{n} \big(1 + 2k \big) \Big) = \frac{1}{n + 1} \Big(1 + n + 2\sum_{k = 1}^{n} k \Big)$$

$$= \frac{1}{n + 1} \Big(1 + n + \frac{2(n+ 1)n}{2} \Big) = \frac{1}{n + 1} + \frac{n}{n + 1} + n = n + 1.$$

In addition, since $\sin \frac{t}{2} > \frac{t}{\pi}$, $\forall n \in \mathbb{N}, \forall t\in (0, \pi]$ we have

$$K_n(t) = \frac{1}{n + 1}\frac{\sin^2(\frac{n + 1}{2}t)}{\sin^2(\frac{t}{2})} \leq \frac{1}{n + 1} \frac{1}{\frac{t^2}{\pi^2}} = \frac{\pi^2}{(n + 1)t^2}.$$

Taking into account the above results it immediately follows that

$$\forall t \in (0, \pi], K_n(t) \leq \min\{n +1, \frac{\pi^2}{(n + 1)t^2}\}$$

which shows the truth of the first statement.

---

Is my work correct for $(1)$? Are there parts that I need to improve? For $(2)$, despite the fact that the function is Lipschitz, I have no idea.

Answer 1

For (2): using that $K_n$ is non-negative and $\int_o^{2\pi}K_n(t)\,dt=1$, we have $$ |\sigma_nf(x)-f(x)|=|K_n\ast f(x)-f(x)|\le\int_0^{2\pi}K_n(t)\,|\,f(x-t)-f(x)\,|\,dt\le K\int_0^{2\pi}K_n(t)\,t\,dt. $$ Using (1) we get $$\begin{align} \int_0^{2\pi}K_n(t)\,t\,dt&\le\int_0^{\pi/(n+1)}(n+1)\,t\,dt+\int_{\pi/(n+1)}^{2\pi}\frac{\pi^2}{(n+1)t}\,dt\\ &<\frac{\pi^2}{2(n+1)}+2\,\pi\,\frac{\log(2(n+1))}{n+1}. \end{align}$$