Let $\xi_1,\xi_2,\dots,\xi_N \in \mathbb{R}$, $p \geq 1$, I want to show
$(\Delta^{(\infty)}(\xi_1,\dots,\xi_N))^{p+1} \leq N(p+1)\Delta^{(p)}(\xi,\dots,\xi_N)$
Definitions :
Let $\psi(x) := \begin{cases}{ x - \lfloor x \rfloor - 1/2 \; x \not\in \mathbb{Z} \\ 0 \; x \in \mathbb{Z} }\end{cases}$
So for istance $\psi \; : \; \mathbb{R} \to \mathbb{R}$ is characterized by the three properties :
- $\psi(x) = x-1/2 \; \text{ if } x \in (0,1)$
- $\psi(n) = 0 \; \text{ for } n \in \mathbb{Z}$
- $\psi(x + 1) = \psi(x) \; \text{ for all } x \in \mathbb{R}$
one defines first the function
$f(y) := {\sum_{n=1}^{N}{\psi(y-\xi_n)}}$
and then defines
$\Delta^{(\infty)}(\xi_1,\dots,\xi_N) := \| f \|_{L^\infty([0,1])}$
$\Delta^{(p)}(\xi,\dots,\xi_N) := \|f\|_{L^p([0,1])}$
My attempt of solution :
First a bit of notation : by an abuse of notation if $f \in L^1(\mathbb{R}/\mathbb{Z})$ and $\mu$ is a signed Radon measure on $\mathbb{R}/\mathbb{Z}$ then I let $\| f\mu \|_{L^1(\mathbb{R}/\mathbb{Z})} := \int_{\mathbb{R}/\mathbb{Z}}{|f(x)|d|\mu|}$ where $|\mu|$ is the variation measure of $\mu$.
my idea is : first $f = \psi * \sum_{n=1}^{N}{\delta(y-\xi_n)}$ so that distributionally $f' = \psi' * \sum_{n=1}^{N}{\delta(y-\xi_n)}$ but $\psi' = 1 - \delta$ so $f'(y) = N - \sum_{n=1}^{N}{\delta(y-\xi_n)}$
now let $g(t) := |t|^{p+1}$
since $\int_{\mathbb{R}/\mathbb{Z}}{f(y)dy} = 0$ one can show that it holds an inequality of the kind
$\|g(f)\|_{L^\infty} \leq \|g'(f)f'\|_{L^1(\mathbb{R}/\mathbb{Z})}$
Now by Holder $\|g'(f)f'\|_{L^1(\mathbb{R}/\mathbb{Z})} = (p+1)\||f|^pf'\|_{L^1(\mathbb{R}/\mathbb{Z})} = (p+1)\int_{\mathbb{R}/\mathbb{Z}}{ |f(x)|^p (Ndx+\sum_{n=1}^{N}{\delta(x-\xi_n)}) } \leq N(p+1)\|f\|^p_{L^p(\mathbb{R}/\mathbb{Z})} +(p+1)\sum_{n=1}^{N}{|f(\xi_n)|^p}$
So at the end I find
$(\Delta^{(\infty)}(\xi_1,\dots,\xi_N))^{p+1} \leq N(p+1)\Delta^{(p)}(\xi,\dots,\xi_N) + (p+1)\sum_{n=1}^{N}{|f(\xi_n)|^p}$
I don't know hot to get rid of the extra piece.