Let $P\equiv P(x) := \sum_{|\alpha|\leq m} c_\alpha\cdot x^\alpha$ be a real polynomial in $d$ variables of (total) degree $m$, where $d, m \in\mathbb{N}$ are fixed.

(I.e., the above sum ranges over all multiindices $\alpha=(i_1, \ldots, i_d)\in\mathbb{N}_0^{\times d}$ of length $|\alpha|\equiv i_1+\ldots + i_d$ less than $m$.)

Denote by $P_k = \sum_{|\alpha|=m}c_\alpha\cdot x^\alpha$, $ 0\leq k\leq m$, the $k$-th homogeneous component of $P$.

I was wondering the following: Given a compact subset $K$ in $\mathbb{R}^d$, is it possible to for $\varphi_K(P):=\max_{0\leq k\leq m}\|P_k\|_{\infty; K}$ (or indeed for any $\ell_p$-norm of $(\|P_k\|_{\infty;K})_{k\geq 0}$ with $1\leq p \leq \infty$) find a constant $\kappa=\kappa(d,K)$ such that

$$\tag{1} \varphi_K(P) \ \leq \ \kappa\cdot \|P\|_{\infty; K} \qquad \text{ for each } \ P \ \text{ as above} \ ?$$

(Here, $\|f\|_{\infty;K}:=\sup_{x\in K}|f(x)|$ is the uniform norm over $K$.) Any references are welcome.

  • $\begingroup$ It looks like in several places you write $m$ where you mean to write $k$. As written the definition of $P_k$ does not depend on $k$. It would also be nice if instead of $||(P_m)||_{\infty;K}$ you just wrote $||(P)||_{\infty;k}$, since the former makes it look like it just depends on the $m$-th homogeneous component. $\endgroup$
    – Vik78
    Commented Dec 18, 2021 at 19:54
  • $\begingroup$ @Vik78 You're right, thanks for pointing that out. I edited accordingly. As for your comment re $(P_k)$, I wrote it that way to indicate that the norms in question refer to (the norms of the components of) the degree-based gradation $(P_k)$ of $P$. $\endgroup$
    – fsp-b
    Commented Dec 18, 2021 at 19:56
  • $\begingroup$ Now you have changed its definition to read $||(P_k)||_{\infty;K}$ with $k$ as a subscript— but you are maximizing over $0 \le k \le m$ so it does not depend on $k$. Personally I find it confusing (and would just replace the entire symbol $||(\cdot)||$ with a function $f(P, K)$, since as written it looks very similar to the norm $||\cdot||$). $\endgroup$
    – Vik78
    Commented Dec 18, 2021 at 20:04
  • $\begingroup$ Do you want to have the constant $\kappa$ independent of the degree of the polynomial? This looks very unlikely. $\endgroup$ Commented Dec 18, 2021 at 21:10
  • $\begingroup$ @JochenWengenroth Yes, $\kappa$ should be independent of the degrees of $P$ and $P_k$. $\endgroup$
    – fsp-b
    Commented Dec 18, 2021 at 21:12

1 Answer 1


The answer is no. E.g., let $d=1$, $K=[0,1]$, and, for $x\in K$, $$P(x):=T_n(x):=n\sum_{0\le k\le n/2}\frac{(-1)^k}{n-k}\binom{n-k}k2^{n-2k-1}x^{n-2k} =\cos(n\arccos x),$$ the $n$th Chebyshev polynomial.

Then $\|P\|_{\infty;K}\le1$, whereas (say) for $k=0$ we have $\|P_k\|_{\infty;K}=2^{n-1}\to\infty$ as $n\to\infty$.


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