Simple yet hard inequality with Lipschitz functions Let $\{w_n\}_{n=1,...N}$ be a sequence of numbers in $\mathbb R$ such that
$$\sum_{i=1}^N w_n = 0\qquad \sum_{i=1}^N |w_n| \le 2.$$
Let $L_1$ denote the class of $1-$Lipschitz continuous functions over $\mathbb R$, i.e. $|f(x)-f(y)|\le |x-y|$ for every $x,y\in \mathbb R$.
Let $H_{\alpha,1}$ denote the class of $\alpha,1-$Lipschitz continuous functions over $\mathbb R$, i.e. $|f(x)-f(y)|\le |x-y|^\alpha$ for every $x,y\in \mathbb R$.
Is it true that, for any other sequence $\{a_n\}_{n=1,...N}$,
$$\sup_{f\in H_{\alpha,1}} \sum_{i=1}^N w_nf(a_n) \le \Big ( \sup_{f\in L_{1}} \sum_{i=1}^N w_nf(a_n) \Big )^\alpha?$$
Note: for $N=2$ the positive answer is very simple, since we can easily find the function realizing the supremum. For higher values of $N$, it seems much harder. I have thinked about using some kind of Jensen's inequality, but nothing forks for negative weights.
 A: ERRATUM. I will try correct soon, since I made an error which invalidates a large part of what is written below. The statement « a function $f$ is $H_{\alpha,1}$ on $[a_1,a_N]$ if and only if its restriction to all sub-intervals $[a_k,a_{k+1}]$ is $H_{\alpha,1}$ » is false, unless $\alpha=1$. I apologize for this mistake.
One needs to put a multiplicative constant (depending on $N$) in the inequality.
I assume that $\alpha \le 1$ and $a_1<\ldots<a_N$. Hence, by sub-additivity of $x \mapsto x^\alpha$ on $\mathbf{R}_+$, a function $f$ is $H_{\alpha,1}$ on $[a_1,a_N]$ if and only if its restriction to all sub-intervals $[a_k,a_{k+1}]$ is $H_{\alpha,1}$.
For $1 \le k \le N$, set
$$W_k := \sum_{i=1}^kw_i.$$
Then, since $W_n=0$, for every $f$ is $H_{\alpha,1}$,
$$\sum_{k=1}^N w_k f(a_k) = \sum_{k=1}^{N-1} W_k \big(f(a_k)-f(a_{k+1})\big),$$
so
$$\Big|\sum_{k=1}^N w_k f(a_k)\Big| \le \sum_{k=1}^{N-1} |W_k| \big|f(a_k)-f(a_{k+1})\big| \le \sum_{k=1}^{N-1} |W_k| \big(a_{k+1}-a_k\big)^\alpha,$$ and for a suitable choice of $f$, these inequalities are equalities.
Hence,
$$\sup_{f \in H_{\alpha,1}} \sum_{k=1}^N w_k f(a_k) = \sup_{f \in H_{\alpha,1}} \Big|\sum_{k=1}^N w_k f(a_k)\Big| = \sum_{k=1}^{N-1} |W_k| \big(a_{k+1}-a_k\big)^\alpha.$$
Of course, the same results holds with $L_1$.
Hence, we are looking for the best constant $C$ such that for every choice of the weights $w_1,\ldots,w_N$ and points $a_1<\ldots<a_N$,
$$\sum_{k=1}^{N-1} |W_k| \big(a_{k+1}-a_k\big)^\alpha \le C\Big(\sum_{k=1}^{N-1} |W_k| \big(a_{k+1}-a_k\big)\Big)^\alpha.$$
By Hölder inequality,
$$\sum_{k=1}^{N-1} |W_k| \big(a_{k+1}-a_k\big)^\alpha = \sum_{k=1}^{N-1} |W_k|^{1-\alpha} |W_k|^\alpha \big(a_{k+1}-a_k\big)^\alpha \le \Big(\sum_{k=1}^{N-1} |W_k|\big)^{1-\alpha} \Big(\sum_{k=1}^{N-1} |W_k| \big(a_{k+1}-a_k\big)\Big)^\alpha,$$
and this is an equality when $|W_k|^{1-\alpha}$ is proportional to $|W_k|^\alpha \big(a_{k+1}-a_k\big)^\alpha$.
Hence, the best constant is
$$C_N := \sup\Big\{\Big(\sum_{k=1}^{N-1} |W_k|\Big)^{1-\alpha} : w_1+\ldots+w_N=0 \textrm{ and } |w_1|+\ldots+|w_N| \le 2\Big\}.$$
When $n \ge 2$, we have $C_N>1$.
My impression is that $C_N$ is achieved when $w_1=1$, $w_N=-1$ and the other $w_k$ are null. If it is true, we would find that $C_N$ is  $(N-1)^{1-\alpha}$.
