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 $t\mapsto\sup_{0\le u\le t}p(u)$.


  • $\Bbb R[X]_0$ 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. $$ ||f||_{\lambda}:=\sup_{0\le s<t\le1}\frac{|f(t)-f(s)|}{|t-s|^{\lambda}}\;\;. $$


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?


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?


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?


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!


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

  • $\begingroup$ In the supremum that defines $\Theta( p)$ the $u$ are nonnegative, aren't they? $\endgroup$ Jul 6, 2017 at 21:29
  • $\begingroup$ @PietroMajer: yes, you're right, I've just edited. Thanks! $\endgroup$
    – Joe
    Jul 7, 2017 at 11:06
  • 1
    $\begingroup$ Is it interesting that $$\Psi(pq) \leq \Psi(p)\Psi(q) \leq \frac{1}{2} (\psi(p^2) + \psi(q^2))? $$ $\endgroup$ Jul 13, 2017 at 10:50
  • $\begingroup$ @IvanDiLiberti: many thanks for your suggestion!! I should work a bit in order to understand if it could be really helpful. But, apart from this, how can you know it is true? Is it trivial (just to write down things and all will be clear), or is there something not trivial? I'm going to use this immediately!! Many thanks again $\endgroup$
    – Joe
    Jul 13, 2017 at 11:17
  • $\begingroup$ Maybe I am wrong but it looks quite obvious to me from definitions and the fact that for all $a,b \in \mathbb{R}^+ ab \leq \frac{1}{2}( a^2 + b^2)$ $\endgroup$ Jul 13, 2017 at 11:20

2 Answers 2


I don't know if this is what you are looking for, but the (essential) supremum of a function can be written as an integral with respect to a (non-additive) measure (namely, the sign of the Lebesgue-measure). This is worked out in Denneberg's Nonadditive Measure and Integral, chapter 9 on Lebesgue spaces. If am not mistaken, you define the set function $\sigma(A) = 0$ if $A$ is a Lebesgue null set and $\sigma(A) = 1$ else. Then the essential supremum of a function is the integral of that function with respect to $\sigma$, i.e. $$ \Psi(p)(t) = \sup_{0\leq u\leq t} p(u) = \int\limits_{[0,t]} p\,d\sigma = \int\limits_{0}^t p\,d\sigma $$ (with the right notion of integral for this non-additive set function $\sigma$). Maybe this helps?

  • $\begingroup$ first of all many thanks +1. Certainly it may helps, I'm gonna give a look on the book you suggested me; for the moment I don't approve your answer only because maybe someone else could have something useful to say. Thanks again. $\endgroup$
    – Joe
    Jul 6, 2017 at 14:58
  • $\begingroup$ I have a question: the $\sup$ is not linear wrt to $f$ while every integral should be it! I mean $\sup(f+g)\neq\sup f+\sup g$ but when we construct a functional, in order to call it "integral" it should preserve certain properties, like linearly: from an integral I expect a behaviour like $\int(f+g)=\int f+\int g$. Maybe this depends on the particular measure we are dealing with but it sounds strange! Don't you think? $\endgroup$
    – Joe
    Jul 13, 2017 at 16:17
  • $\begingroup$ The integral (in the sense Denneberg defines it) for the nonadditive measure is indeed not linear. So, no, not every integral should be linear $\endgroup$
    – Dirk
    Jul 13, 2017 at 16:19

Equivalently you can also define $\theta:=\Psi( p)$ as the smallest increasing function on $I:=[0,1]$ which is larger than $p$ on $I$ (then your definition would rather be the construction of $\theta$). Also note that you can define this inferior envelope of increasing functions for any bounded function $p$ on $I$.

A simple fact: by the construction, if $p$ is continuous, for any $0\le s \le t\le 1$ with $\theta(s)\neq \theta(t)$ there are $ s\le s'\le t'\le t $ such that $p(t')=\theta(t)$ and $p(s')=\theta(s) $. As a consequence, if $\omega$ is a(n increasing) modulus of continuity for $p$, then $\omega$ is also a modulus of continuity for $\theta$. In particular, if $p$ is $k$-Lipschitz so is $\theta$, and if $p$ is $\lambda$-Hölder, so is $\theta$, with $\|\theta\|_\lambda\le\|p\|_\lambda$ .

(For a polynomial $p$, one can also observe that $\theta$ is a piecewise $C^1$ function, hence Lipschitz. In fact $\|\theta'\|_{\infty,[0,1]}\le\| p'\|_{\infty,[0,1]}$, because either $t$ is a local maximum of $p$, or $\theta'(t)=0$ or $\theta'(t)=p'(t)$, so $\rm{Lip}(\theta)\le \rm{Lip}( p)$).

rmk. As to examples of uses in the literature and properties, I'd say this function $\theta$ more or less explicitly appears in F.Riesz' Running water lemma (a key ingrediend of his proof of Birkhoff's individual ergodic theorem). In fact, I believe the name "running water" alludes to the section profile of a hill (the graph of $p$) when water is streaming from the right and fills the holes (the graph of $\theta$).

  • $\begingroup$ Did you mean that there exist $s',t'$ with $0\leq s'\leq t\leq t'\leq 1$ such that $p(t') = \theta(t)$ and $p(s') = \theta(s)$. Otherwise I don't see how these $s'$ and $t'$ should exist, and also this gives the right estimate $(\theta(t)-\theta(s))/(t-s) = (p(t')-p(s'))(t-s) \leq (p(t')-p(s'))/(t'-s')$. $\endgroup$
    – Dirk
    Jul 12, 2017 at 8:06
  • $\begingroup$ Right, I meant that for a continuous $p$, and I should have added: for any $0\le s \le t\le1$ such that $\theta(s)\neq \theta(s)$ (edited). In this case the existence of $s\le s'\le t'\le t$ as required is OK (and it is sufficient to conclude). There are of course other statements that do; yours seems simpler and more general. $\endgroup$ Jul 13, 2017 at 15:00

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