**THE FRAMEWORK**

Let $0<\lambda\le1$ and consider $$ \Psi:(\Bbb R[X]_0,||\cdot||_{\lambda})\longrightarrow(\mathcal C^{\lambda}[0,1],||\cdot||_{\lambda}) $$ defined as $$ \Psi(p):=\sup_{0\le u\le\cdot}p(u) $$ in the sense that the polynomial $p$ is sent by $\Psi$ to the function $t\mapsto\sup_{0\le u\le t}p(u)$.

Here:

$\Bbb R[X]_0$ denotes the space of one variable polynomials with real coefficients which vanish at $0$;

$\mathcal C^{\lambda}[0,1]$ is the space of $\lambda-$ Holder continuous functions $f:[0,1]\to\Bbb R$;

- $||\cdot||_{\lambda}$ denotes the usual $\lambda-$Holder
*seminorm*, i.e. $$ ||f||_{\lambda}:=\sup_{0\le s<t\le1}\frac{|f(t)-f(s)|}{|t-s|^{\lambda}}\;\;. $$

**THE PROBLEM**

What kind of function is $\Psi$? What property does it have? Does $\Psi$ belong to a family of functions which has a name and for which a theory is developed somewhere? Can someone suggest me some reference?

But the main problem for me is: is there a way to prove/disprove $\Psi$ is Lipschitz?

**APPROACH 1: LAGRANGE**

After various approaches, I think a way could be the following: let's start with $\lambda=1$; then I think that for every $0\le s<t\le1$, we can find $0\le s^*\neq t^*\le1$ s.t. $$ \frac{\Psi(p)(t)-\Psi(p)(s)}{t-s}= \frac{p(t^*)-p(s^*)}{t^*-s^*} $$ and by Lagrange, the last one equals $p'(u)$, for some $u\in]s^*,t^*[$ (supposing wlog that $s^*< t^*$).

Thus, if my previous conjecture was true, for every $0\le s<t\le1$, we can find $u,v\in]0,1[$ such that \begin{align} \frac{\Psi(p)(t)-\Psi(p)(s)-(\Psi(q)(t)-\Psi(q)(s))}{t-s}=p'(u)-q'(v) \end{align} from which \begin{align*} ||\Psi(p)-\Psi(q)||_1 &=\sup_{0\le s<t\le1} \frac{|\Psi(p)(t)-\Psi(p)(s)-(\Psi(q)(t)-\Psi(q)(s))|}{|t-s|}\\ &=\sup_{u,v}|p'(u)-q'(v)|\\ &\le\sup_{0\le a<b\le1\\0\le c<d\le1}\left|\frac{p(b)-p(a)}{b-a}-\frac{q(d)-q(c)}{d-c}\right| \end{align*} Now, from this, it's clear I've lost something, since the last term is clearly $\ge$ than $$ ||p-q||_1= \sup_{0\le s<t\le1} \frac{|p(t)-p(s)-(q(t)-q(s))|}{|t-s|}\ $$ but I think this could be the right way; going further in this direction, I noticed the problem can be solved if I prove the following proposition:

Fix $p,q\in\Bbb R[X]_0$; then for every $0\le s< t\le1$ we have defined $u=u(s,t)$ and $v=v(s,t)$. So, for every such $u,v\in]0,1[$ there exists $\alpha\in]0,1[$ such that $$ |p'(u)-q'(v)|=|p'(\alpha)-q'(\alpha)| $$

Does someone have any suggestion? Any hint? Any feeling about the truthfulness of the fact that $\Psi$ is Lipschitz?

**APPROACH 2: SLOPES**

Working on this problem I thought that a "pointwise approach" could be too strong to attack the problem. Maybe a useful point of view could be to see the already written ratio
$$
\frac{\Psi(p)(t)-\Psi(p)(s)-(\Psi(q)(t)-\Psi(q)(s))}{t-s}
$$
as a *difference of slopes*:
$$
\frac{\Psi(p)(t)-\Psi(p)(s)}{t-s}-\frac{\Psi(q)(t)-\Psi(q)(s)}{t-s}
$$
and noticing that the incriminated object $||\Psi(p)-\Psi(q)||_1$ is the supremum of the absolute value of the difference of these slopes.

Now, let's denote, for $0\le s<t\le1$ $$ \Delta_{s,t}\Psi(p):=\frac{\Psi(p)(t)-\Psi(p)(s)}{t-s} $$ it's clear that $\Delta_{s,t}\Psi(p)\ge0$ for every $s,t,p$.

How can we prove that the differences of the slopes of $\Psi(p),\Psi(q)$ (which are $|\Delta_{s,t}\Psi(p)-\Delta_{s,t}\Psi(q)|$) is $\le$ of the differences of slopes of $p,q$,

not pointwise, but passing at $\sup$?

Here again: there exists some book/paper which treat slopes in a way which can be useful in this context?

**APPROACH 3: RELATED POLYNOMIALS**

I noticed that for every couple of $p,q\in\Bbb R[X]_0$, we can find $\widetilde p,\widetilde q\in\Bbb R[X]_0$ strictly increasing, such that $$ ||\Psi(p)-\Psi(q)||_1 \le||\Psi(\widetilde p)-\Psi(\widetilde q)||_1 =||\widetilde p-\widetilde q||_1 $$ but the problem is that a priori we don't know how the last term written behaves with respect to $||p-q||_1$; but I thought: is there a way, given $p,q\in\Bbb R[X]_0$ to find $\widetilde p,\widetilde q\in\Bbb R[X]_0$ which are "more maneuverable" than $p,q$ and allow to get achieve some useful result?

I know that it is vague, but this was only an idea and obviously I'm going to think about it, but at the same time I'm trying to update here every idea which seems to be meaningful!

**CONTEREXAMPLE**

After days of efforts trying to prove $\Psi$ is Lipschitz, I tried to search a counterexample.

Here it is: $$ p(x)=x^n\\ q(x)=x^n-x=x(x^{n-1}-1) $$ it's clear that $\Psi(p)\equiv p$ while $\Psi(q)\equiv 0$ (remember we are working on $[0,1]$). In this way we have \begin{align*} ||\Psi(p)-\Psi(q)||_1&=||p||_1=p'(1)=n\\ ||p-q||_1&=1 \end{align*}

Do you think it is correct?

Many thanks