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Let $f: \mathbb{T} \to \mathbb{R}$ with $m \leq f(x) \leq M$ for some $m, M \in \mathbb{R}$ and all $x \in \mathbb{T}$. We have, the $k$-th Cesàro mean of the Fourier series $f$: $$\sigma_k[f](x) = \int_{-\pi}^{\pi}F_k(t)f(x-t)\ dt,$$ where $F_k(t)$ is the Fejér kernel.

To show: $m \leq \sigma_k[f] \leq M$ for all $x \in \mathbb{T}$ and $k \in \mathbb{N}$.

I thought because $m \leq f(x) \leq M$ , then $m \leq f(x-t) \leq M$, by the periodicity of $f$. Using $\int_{-\pi}^{\pi}F_k(t)\ dt = 1$, it is clear that $m \leq \sigma_k[f] \leq M$. Then I'm starting to doubt whether I can take such conclusion. I think this reasoning will work if $f(x) \geq 0$ for all $x \in \mathbb{T}$. Any comments or tips would be appreciated.

Vicky
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1 Answers1

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Hint: Note that Fejér kernel is an approximate to identity, and $$F_k(t)=\frac{1}{k+1}\left(\frac{\sin\frac{(k+1)t}{2}}{\sin\frac{t}{2}}\right)^2\geq 0.$$ So, $$m=\frac{m}{2\pi}\int_{-\pi}^\pi F_k(t)\ dt\leq\sigma_k(f)(x)=\frac{1}{2\pi}\int_{-\pi}^\pi F_k(t) f(x-t)\ dt\leq \frac{M}{2\pi}\int_{-\pi}^\pi F_k(t)\ dt= M.$$

Note that there is another alternative formulation of Fejér kernel given by $$F_k(t)=\sum_{\ell=-k}^k\left (1-\frac{|\ell|}{k+1}\right)e^{i\ell\theta}.$$ But, we do not need this here.

Exodd
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Sumanta
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  • Maybe you mean $m = m\int_{-\pi}^{\pi}F_k(t)\ dt$? Since the integral of the Fejér kernel is 1, right? – Vicky Sep 23 '20 at 11:18
  • Yes, the integral of the Fejér kernel is $1$, some use the factor $\frac{1}{2\pi}$, some not. It depends on the definition of Fejér kernel, in order to make it an approximate to identity we have to multiply it by suitable constant. – Sumanta Sep 23 '20 at 11:19
  • So we don't have to worry that $f(x)$ could be negative? – Vicky Sep 23 '20 at 11:21
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    No, you don't need to consider this special case as Fejér kernel is non-negative, as I have written. – Sumanta Sep 23 '20 at 11:22
  • What do you mean with "it's an approximate to identity"? – rreevv97 Apr 28 '21 at 06:36
  • Besides, how do you conclude those inequalities? – rreevv97 Apr 28 '21 at 06:39