Can a norm on polynomials be supermultiplicative? A norm on a real algebra is supermultiplicative when $\lVert f\cdot g\rVert\geq\lVert f\rVert\cdot\lVert g\rVert$ for all $f$ and $g$ in the algebra.
Is there a supermultiplicative norm on $\mathbb R[x]$?
This is a one-sided form of my previous question. An answer to either question could provide an answer to the other.
I suspect that the norm given by $\Big\lVert\sum_ka_kx^k\Big\rVert=\max_k(k!|a_k|)$, or equivalently by $\Big\lVert\sum_ka_kx^k/k!\Big\rVert=\max_k|a_k|$, is a multiple of a supermultiplicative norm (so $\lVert f\cdot g\rVert\geq C\lVert f\rVert\lVert g\rVert$ for some constant $C>0$). Is this true?
 A: (1) Yes, there are supermultiplicative norms. (2) No, no positive multiple of the norm defined by $\Big\lVert\sum_ka_kx^k\Big\rVert=\max_k(k!|a_k|)$ is supermultiplicative.
Proofs:
(1) We’ll construct a norm of the form $\|\sum a_kx^k\|=\max_k(c_k|a_k|).$ We construct the $c_k$ to ensure supermultiplicativity, with a slightly stronger property to help the induction.
Take $c_0=c_1=\tfrac 1 2.$ For each $n\geq 1,$ suppose for induction that we have constructed $c_0,\dots,c_n>0$ such that, if we define a seminorm by $\|\sum a_kx^k\|_n=\max_{k\leq n}(c_k|a_k|),$ then
$$\|fg\|_n\geq (1+2^{-\deg(fg)})\|f\|_n\cdot\|g\|_n\tag{†}$$
holds whenever $0\leq \deg(fg)\leq n.$
Consider the set $S$ of pairs of polynomials $(f,g)$ such that: $\deg(f),\deg(g)\leq n,$ and $\deg(fg)\leq n+1,$ and $\|f\|_n=\|g\|_n=1,$ and $\|fg\|_n\leq 1+2^{-(n+1)}.$ The set $S$ is compact (with the “usual” topology, as a subspace of a finite dimensional vector space). Let $m$ be the minimum absolute value of the coefficient of $x^{n+1}$ in $f(x)g(x),$ over all $(f,g)\in S.$ This is attained by compactness of $S.$ We must have $m\neq 0,$ because $m=0$ would imply (†) which contradicts $(f,g)\in S.$ Taking $c_{n+1}=(1+2^{-(n+1)})/m$ works: (†) holds with $n$ replaced by $n+1.$ (We only needed to check factors of degree $\leq n$ because (†) holds when $f$ or $g$ is constant and $c_0\leq \tfrac 1 2.$)
(2) Consider $f_N(x)=\prod_{n=1}^N(1-x/\pi n)$ and $g_N(x)=xf_N(-x)=x\prod_{n=1}^N(1+x/\pi n).$ The $x$ coefficient of $f_N(x)$ tends to $-\infty$ as $N\to\infty,$ and similarly the $x^2$ coefficient of $g_N(x)$ tends to $\infty.$ So $\|f_N\|\cdot\|g_N\|$ is unbounded.
But $\|f_Ng_N\|\leq 1.$ Let $a_{N,k}$ be the $x^k$ coefficient of $f_N(x)g_N(x)=x \prod_{n=1}^N(1-x^2/\pi^2 n^2).$ If we expand the product, all the terms contributing to $a_{N,k}$ have the same sign: there are no terms for even $k$ and the sign for odd $k$ is the same as $(-1)^{(k-1)/2}.$ So $|a_{N,k}|$ converges from below.
The sequence $f_Ng_N$ converges locally uniformly to $\sin x$ - this is a standard exercise with the Hadamard factorization thorem. Locally uniform convergence gives convergence of Taylor series coefficients by Cauchy’s differentiation formula. Hence $|a_{N,k}|\leq 1/k!$ as required.
