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Consider the space $V_{d,n}=\mathbb{R}[x_1,\ldots,x_n]_d$ of homogeneous polynomials of degree $d$ in $n$ variables. I am interested in the set of bounded polynomials on the sphere $$B_{d,n}=\{f\in V_{d,n}:\, ‖f(x)‖\leq 1\textrm{ for all } ‖x‖=1\}.$$ Assume that we have fixed $d$ and $n$. I am looking for a "universal polynomial" for the set $B_{d,n}$. By that I mean a polynomial $F\in B_{d,N}$ for some possibly large $N$ such that every element $f\in B_{d,n}$ can be obtained by restricting $F$ to some suitable linear subspace $V\subset\mathbb{R}^N$ with $\dim(V)=n$. To be more precise: For each $f\in B_{d,n}$ there should be a norm preserving linear map $U:\mathbb{R}^n\to\mathbb{R}^N$ such that $f=F\circ U$.

I can find such a polynomial for $d=1,2$ (see below) and I would be very interested in the case $d=3$:

Question. Is there such a universal polynomial for $d=3$ and any $n$?

I would highly appreciate any answer to that question, positive or negative.

Sketch for $d=1,2$: Here I want to sketch the construction of such a polynomial for $d=1,2$. For $d=1$ we claim that $N=n+1$ and $F=x_{n+1}$ does the job. Indeed, if $f\in B_{1,n}$, then after an orthogonal change of coordinates, we can assume that $f=\lambda x_1$ with $|\lambda|\leq1$. Then there is a $t\in\mathbb{R}$ such that $\lambda=\cos(t)$. Then we can apply the following orthogonal change of coordinates to $F$:

$$x_{n+1}\mapsto \cos(t)x_1+\sin(t) x_{n+1},\,\, x_1\mapsto -sin(t)x_1+\cos(t)x_{n+1}$$ and we obtain $\cos(t)x_1+\sin(t) x_{n+1}$. Restricting this to the $n$-dimensional subspace $x_{n+1}=0$ we obtain $f$. This shows the case $d=1$.

For the case $d=2$ we claim that $N=3m$ and the polynomial $F=x_{n+1}^2+\ldots+x_{2n}^2-(x_{2n+1}^2+\ldots+x_{3n}^2)$ does the job. To see that we note that any $f\in B_{2,n}$ can be written as $\lambda_1x_1^2+\ldots+\lambda_nx_n^2$, $|\lambda_i|\leq1$, after some orthogonal change of coordinates. Then we can apply orhogonal transformations to $F$ similar as in the case $d=1$ for each $x_i$, $1\leq i\leq n$, and obtain $f$ by restricting to $x_j=0$ for $j>n$.

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    $\begingroup$ "such that every element $f\in B_{d,n}$ can be obtained by restricting $F$ to some suitable linear subspace" Just to clarify what exactly you mean by this. Is it that there exists a norm-preserving linear mapping $U:\mathbb R^n\to\mathbb R^N$ such that $f=F\circ U$, or is it something stronger? $\endgroup$
    – fedja
    Commented Aug 31, 2020 at 16:21
  • $\begingroup$ Yes, this is what I mean. Thanks, I will also edit the question accordingly. $\endgroup$
    – Hans
    Commented Aug 31, 2020 at 16:23
  • $\begingroup$ Is your $F$ for $d=1$ bounded by 1? $\endgroup$ Commented Aug 31, 2020 at 20:22
  • $\begingroup$ @FedorPetrov As written, obviously not, but $x_{n+1}$ alone would do just fine. $\endgroup$
    – fedja
    Commented Sep 1, 2020 at 0:11
  • $\begingroup$ Yes, thanks. I adjusted the case $d=1$ accordingly. $\endgroup$
    – Hans
    Commented Sep 1, 2020 at 9:02

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