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

Is it possible to estimate the maximum coefficient $\|c_\alpha\|_\infty := \max_{|\alpha|\leq m}|c_\alpha|$ of $P$ against its uniform norm $\|P\|_{\infty;K}:= \sup_{x\in K}|P(x)|$, for $K$ some compact set in $\mathbb{R}^d$?

That is, does there exist a compact set $K$ in $\mathbb{R}^d$ together with a constant $\kappa \equiv \kappa(m,d,K)>0$ such that

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

Any references are welcome.

Edit: Do you know if $\kappa$ can be chosen independent of $m$, or at least such that the sequence $(\kappa(m,d,K))_{m\geq 0}$ is bounded (for $K$ and $d$ fixed)?

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  • $\begingroup$ is it okay if $K$ depends on $m$ and $d$? $\endgroup$
    – Squala
    Commented Dec 16, 2021 at 17:13
  • $\begingroup$ $K$ is allowed to depend on $d$, but should not depend on $m$. $\endgroup$
    – fsp-b
    Commented Dec 16, 2021 at 17:15
  • $\begingroup$ Any two norms on a finite-dimensional space are equivalent. $\endgroup$ Commented Dec 16, 2021 at 17:41
  • $\begingroup$ Hi rmceraf and welcome to the MathOverflow: though it is not forbidden to cross post across the site and the Mathematics Stack Exchange, 1. At least a week should have passed between the two post, and 2. It should be stated in the body of the question. These requirements are just to let all users the possibility to answer and avoid cluttering the two web sites. $\endgroup$ Commented Dec 16, 2021 at 20:29
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    $\begingroup$ Thanks, Daniele. Does anyone have an idea if independence (of $\kappa$) of $m$ could also be achieved? $\endgroup$
    – fsp-b
    Commented Dec 17, 2021 at 4:45

2 Answers 2

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Yes, such a $\kappa$ exists for every compact set $K$ with non-empty interior. Here is an abstract linear-algebra argument.

Let $V$ be the real vector space spanned by the multi-indices $\alpha$ with length at most $m$. We have a linear map $A\colon V \to \mathbb R^K$, sending $(c_\alpha)$ to the function $x\to \sum_\alpha c_\alpha x^\alpha$. Since different polynomials cannot agree on a set with non-empty interior, this map is injective. There is then a finite subset $F\subset K$ such that the composition $V\xrightarrow A\mathbb R^K\to\mathbb R^F$ is a linear isomorphism (for example, pick the elements of $F$ one by one, making sure that the rank of the composition increases by 1 each time).

Now $\kappa$ exists by the fact that all linear isomorphisms between finite-dimensional vector spaces are bicontinuous.

If you want an explicit value for $\kappa$, you can use discrete derivatives (see https://en.wikipedia.org/wiki/Finite_difference) to write $c_\alpha$ as a linear combination of some values of $P$.

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  • $\begingroup$ This constant depends on $m$, as I ubderstood comments to OP it should not $\endgroup$ Commented Dec 16, 2021 at 18:13
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    $\begingroup$ The comment just states $K$ should not depend on $m$. But $\kappa$ can (this is clear from the notation $\kappa = \kappa(m,d,K)$.) $\endgroup$
    – Squala
    Commented Dec 16, 2021 at 18:24
  • $\begingroup$ ah, yes, you are right $\endgroup$ Commented Dec 16, 2021 at 18:46
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    $\begingroup$ One can also use Bernstein's inequality en.wikipedia.org/wiki/Bernstein%27s_theorem_(polynomials) to get reasonable bounds, particularly in low dimensions. $\endgroup$
    – Terry Tao
    Commented Dec 16, 2021 at 22:53
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    $\begingroup$ no, for example if $d=1$ and $K=[0,1]$ the Bernstein polynomials en.wikipedia.org/wiki/Bernstein_polynomial are bounded on $K$ but the coefficients get large $\endgroup$
    – Squala
    Commented Dec 17, 2021 at 9:43
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I've recently been looking into this problem. Here's what I've came up with to show the existence of such a $\kappa$. Let me know if there are any issues. (I gave the same answer on the cross post)

Define the following polynomial space $$ \mathcal{P}_{m,d,K} := \left\{ P: P(x) = \sum_{\alpha:|\alpha|\leq m} c_\alpha x^\alpha, \quad x\in K, \;c_\alpha \in \mathbb{R} \; \forall \alpha \right\} . $$ It can be shown that $\mathcal{P}_{m,d,K}$ is a vector space. Further, since every $P\in\mathcal{P}_{m,d,K}$ can be represented as a linear combination of finitely many basis functions ($x^\alpha$), then $\mathcal{P}_{m,d,K}$ is a finite dimensional vector space.

Then we have the following two polynomial norms, which are both valid norms on $\mathcal{P}_{m,d,K}$: $$ \lVert P \rVert_{\infty;\,K} := \sup_{x\in K}|P(x)| = \max_{x\in K}|P(x)|, $$ and $$ \lVert P \rVert_{\infty;\,c_\alpha} := \max_{\alpha:|\alpha|\leq m}|c_\alpha| = \lVert c_\alpha \rVert_\infty. $$ It is a well known result that all norms on finite dimensional vector spaces are equivalent (Link1, Link2). Therefore, from the definition of equivalent norms, there exists a $\kappa>0$ such that
$$ \lVert P \rVert_{\infty;\,c_\alpha} \leq \kappa\lVert P \rVert_{\infty;\,K} . $$

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  • $\begingroup$ Thanks for your nice answer. Your approach seems correct to me. $\endgroup$
    – fsp-b
    Commented May 24, 2023 at 23:18

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