Operator on Polynomials I have a question. At the moment, I have a (linear) operator $T:\mathcal{P} \to \mathcal{P}$, where $\mathcal{P}$ are the polynomials in $n$-real variables, with the following two properties:

*

*$T$ maps polynomials of total degree $m$ to polynomials of total degree $m$ (for all $m$).

*The exists a constant $C\ge 1$ with
$$||Tf||_{\infty,B_r} \le C^{\mathrm{deg} f} \cdot ||f||_{\infty,B_r},$$
where $B_r$ is the closed ball with respect to the supremums norm on $\mathbb{R}^n$ and $||\cdot ||_{\infty, B_r}$ is the supremums norm on the ball $B_r$.

I would like to know, wether there exist constants $r', D\ge 0$ with
$$||Tf||_{\infty,B_r}\le D \cdot ||f||_{\infty, B_{r'}}, \; \mathrm{ for } \, \mathrm{ all }\, f \in \mathcal{P}.$$
My problem is if it is not the case, I can't find an counterexample, since the statement is correct for all homogeneous polynomials by choosing $D=1, r'=Cr$...
May you could help me?
Best regards,
Dominik
 A: The answer is negative, even for $n=1$. Consider this case, and $T : P \mapsto P'$ the differentiating operator. First, this operator does satisfy your constraint with $r=1$ (and for all $r$, actually): Markov's inequality tells us that for all $P \in \mathbb{R}_n[X]$, $$||P'||_{\infty, [-1,1]} \le n^2 ||P||_{\infty, [-1,1]}$$
with the bound being attained for the Chebyshev polynomial $T_n$. You can take $C=3$ and use that $n^2 \le 3^n$ .

Assume that you have $D$ and $r'>1$ such that for all $P \in \mathbb{R}[X]$, $||P'||_{\infty, [-1,1]} \le D ||P||_{\infty, [-r',r']}$. Drawing inspiration from the previous case, we look at what happens for Tchebychev polynomials scaled to $[-r', r']$. For all $n$, define $S_n(X) = T_n\Big(\frac{X}{r'}\big)$, which is such that for $x \in [-r', r']$, $S_n(x) = \cos\Big(n \arccos\big(\frac{x}{r'}\big)\Big)$. Then $||S_n||_{\infty, [-r',r']} = 1$. Meanwhile, since we took $r'>1$,the derivative can be expressed very simply on $[-1,1]$:
$$|S_n'(x)| = \Big|\frac{n\sin\big(n \arccos\big(\frac{x}{r'}\big)\big)}{r'\sqrt{1-\big(\frac{x}{r'}\big)^2}}\Big| \ge \frac{n}{r'} \cdot \big|\sin \big(n \arccos\big(\frac{x}{r'}\big)\big)\big|$$
For $n$ large enough, the range of $x \mapsto n \arccos\big(\frac{x}{r'}\big)$ on $[-1,1]$ is an interval with length at least $\pi$, so it includes an element of the form $k\pi + \frac{\pi}{2}$, and thus for $n$ large enough, $||S_n'||_{\infty, [-1,1]} \ge \frac{n}{r'}$. This directly contradicts that we should have $||S_n'||_{\infty,[-1,1]} \le D$.
