Is the operator $\; \Theta:\Bbb R[X]\to\mathcal C^{\lambda}[0,1]$ Lipschitz?

Let $$\Theta:(\Bbb R[X],||\cdot||_{\lambda})\longrightarrow(\mathcal C^{\lambda}[0,1],||\cdot||_{\lambda})$$ defined as $$\Theta(p):=\sup_{u\le\cdot}p(u)$$ where $||\cdot||_{\lambda}$ denotes the usual $\lambda-$Holder norm, with $0<\lambda\le1$, i.e. $$||f||_{\lambda}:=||f||_{\infty}+\sup_{0\le s<t\le1}\frac{|f(t)-f(s)|}{|s-t|^{\lambda}}$$ and $\mathcal C^{\lambda}[0,1]:=\{f:[0,1]\to\Bbb R\;:\;||f||_{\lambda}<+\infty\}$.

Is $\Theta$ a Lipschitz operator? Is there a quick way to prove/disprove my statement?

If $\Theta$ were linear it would suffice to show the lipschitzianity near the zero polynomial; but $\Theta$ is sublinear: what can be said about these operators?

EDIT: the answer is NO: just take $p(u)=1+u$ and $q(u)=-u^2$.

However my original problem was less general than this: I have to prove (or disprove... but I hope it's true!) that $$||\sup_{u\le\cdot}p(u)-\sup_{u\le\cdot}q(u)||_{\lambda}\le C||p-q||_{\lambda}$$ for all $p,q$ real polynomial of the same degree. For this it suffices to show the following $$\max_{t\in[0,1]}\left|\sup_{u\le t}p(u)-\sup_{u\le t}q(u)\right|\le \max_{t\in[0,1]}\left|p(t)-q(t)\right|\\ \left|\sup_{u\le t}p(u)-\sup_{u\le t}q(u)-\sup_{u\le s}p(u)+\sup_{u\le s}q(u)\right|\le\left|p(t)-q(t)-p(s)+q(s)\right|\;\;\forall\;\; 0\le s<t\le1$$ but after this I migrated to MO, it seemed more suitable for this problem; but if someone has any hints, guys, you're welcome!

• What is the definition of $\| p \|_\lambda$ for $p$ a polynomial? I assume you mean to take the Hoelder norm of the corresponding function on $[0,1]$? Jul 4, 2017 at 13:58
• I mean exactly this.
– Joe
Jul 4, 2017 at 13:59
• $\max_{t\in[0,1]}\left|\sup_{u≤t} p(u) - \sup_{u≤t}q(u)\right|≤\max_{t\in[0,1]}|p(t)-q(t)|$ is not true unless you are looking at $\sup_{u≤t}|p(u)|$. Let $p$ be a parabola that starts at zero, goes to some negative minimum at $1/2$ and comes back to zero at $1$. Let $q$ be a parabola that starts at a small negative value and then goes to $0$ at $1/2$ and then back to the small negative value. The first expression is $|q(0)|$, the second is $|p(1/2)|$. Jul 4, 2017 at 15:57

I have found a counterexample: just take $p(x)=x^n$ and $q(x)=x^n-x$; in this way, setting $\lambda=1$ we would have $\Theta(p)\equiv p$ and $\Theta(q)\equiv0$ and thus $$\|\Theta(p)-\Theta(q)\|_1=\|p\|_1=n+1$$ and $$\|p-q\|_1=\|x\|_1=2$$ So $\Theta$ can't be Lipschitz.