Can a norm on polynomials be "almost multiplicative", even for large degrees? Definition: A norm on a real algebra is called almost multiplicative if there are positive constants $L$ and $U$ such that, for all $f$ and $g$ in the algebra,
$$L\lVert f\rVert\cdot\lVert g\rVert\;\leq\;\lVert f\cdot g\rVert\;\leq\;U\lVert f\rVert\cdot\lVert g\rVert;$$
equivalently, if
$$0\;<\;\inf_{f,g}\frac{\lVert f\cdot g\rVert}{\lVert f\rVert\cdot\lVert g\rVert}\;\leq\;\sup_{f,g}\frac{\lVert f\cdot g\rVert}{\lVert f\rVert\cdot\lVert g\rVert}\;<\;\infty.$$
Is there an almost multiplicative norm on $\mathbb R[x]$?
The infimum (or supremum) is the same, whether the polynomials $f$ and $g$ have integer coefficients, or rational coefficients (just factor out a common denominator), or real coefficients (take limits of rationals). I don't know if it's the same for complex coefficients (given that the norm respects the complex absolute value: $\lVert e^{i\theta}f\rVert=\lVert f\rVert$).

Denote by $\mathbb R[x]_n$ the space of polynomials of degree $n$ or less. This has finite dimension $n+1$.
If we consider multiplication $\mathbb R[x]_m\times\mathbb R[x]_n\to\mathbb R[x]_{m+n}$, and take the infimum/supremum over these subspaces, then any norm is almost multiplicative. That is because the unit sphere in finite dimensions is compact; the infimum is actually a minimum, and the supremum is actually a maximum. The minimizing polynomials are non-zero, so their product is non-zero: $\frac{\lVert f\cdot g\rVert}{\lVert f\rVert\cdot\lVert g\rVert}>0$. The maximizing polynomials have a defined product, which has finite norm: $\frac{\lVert f\cdot g\rVert}{\lVert f\rVert\cdot\lVert g\rVert}<\infty$.
Of course that argument doesn't work for the whole infinite-dimensional space $\mathbb R[x]$.
No norm on polynomials is actually multiplicative; the infimum is not equal to the supremum. Multiplicative norm on $\mathbb{R}[X]$.
These bounds on norms of products are useful for factoring polynomials with integer coefficients. Suppose $f\cdot g\in\mathbb Z[x]_n\backslash\{0\}$ is known but $f$ and $g$ are not known. Let $L$ be a lower bound for products as described above. Let $\varepsilon>0$ be the minimum norm on $\mathbb Z[x]_n\backslash\{0\}$; this exists because $\mathbb Z[x]_n$ is a lattice in $\mathbb R[x]_n$, and a lattice intersected with a ball is a finite set. Then there are only finitely many possible factors $f$, because
$$\lVert f\rVert\;\leq\;\frac{\lVert f\rVert\cdot\lVert g\rVert}{\varepsilon}\;\leq\;\frac{\lVert f\cdot g\rVert}{\varepsilon L}.$$
Relevant links:
Bombieri Norm
Bounds on Factors in $\mathbb Z[x]$ (Abbott)
Global optimization: a model problem, or A Model Problem for Global optimization (Rump)

I considered weighted $\infty$-norms; norms of the form
$$\Bigg\lVert\sum_ka_kx^k\Bigg\rVert_c=\max_k(c_k|a_k|)$$
where $c=(c_0,c_1,c_2,\cdots)$ is a sequence of positive numbers. The upper bound $\lVert f\cdot g\rVert\leq U\lVert f\rVert\lVert g\rVert$ then is saying that
$$\max_l\left|c_l\sum_{j+k=l}a_jb_k\right|\leq U\max_j|c_ja_j|\max_k|c_kb_k|.$$
By scaling the coefficients, $a_j'=c_ja_j$ and $b_k'=c_kb_k$, this is equivalent to
$$\max_l\left|\sum_{j+k=l}\frac{c_l}{c_jc_k}a_j'b_k'\right|\leq U\max_j|a_j'|\max_k|b_k'|.$$
By applying the triangle inequality (for one direction, and by taking particular examples with $a_j',b_k'\in\{0,1\}$ for the other direction), this is equivalent to
$$\sum_{j+k=l}\frac{c_l}{c_jc_k}\leq U$$
for every $l\in\mathbb N$.
For the constant weight $c_k=1$, the $l$'th sum here is just $\sum_{j+k=l}(1)=l+1$ which is unbounded.
For the factorial weights $c_k=k!$, the $l$'th sum is
$$\sum_{j+k=l}\frac{l!}{j!k!}=\sum_k\binom{l}{k}=2^l$$
which is also unbounded.
For the inverse factorial weights $c_k=1/k!$, the $l$'th sum is $\sum_k1/\binom lk$, which has maximum $8/3$ (at $l=3$ or $4$) and limit $2$ (as $l\to\infty$). So we can take $U=8/3$ as an upper bound. But I think there is no lower bound $L>0$, considering for example $f_n(x)=(x+1)^n$ and $g_n(x)=(x-1)^n$ (though I'm not sure the norm ratio actually converges to $0$ as $n\to\infty$).
For the square weights $c_k=(k+1)^2$, the $l$'th sum is $\sum_k\left(\frac{(l+1)}{(l-k+1)(k+1)}\right)^2$, which seems to have maximum $3.5171$ (at $l=19$), and has limit $\pi^2/3$. So there is an upper bound.
It is harder to find lower bounds. I don't know if there's any norm on $\mathbb R[x]$ for which $\frac{\lVert f\cdot g\rVert}{\lVert f\rVert\cdot\lVert g\rVert}\geq L>0$. Even if there are norms with lower bounds as well as norms with upper bounds, there isn't necessarily a single norm which is bounded both ways.
 A: No, there is no almost multiplicative norm on $\mathbb R[X].$ The argument is an adaptation of the argument for multiplicative norms, using the following statement (copy and pasted from https://math.stackexchange.com/a/2674259/3643)

Theorem (Gelfand–Mazur). Let $A$ be a real normed division algebra. Then $A$ is isomorphic with $\mathbb{R}$, $\mathbb{C}$, or the algebra of quaternions $\mathbb{H}$.
For a proof, see theorem 14.7 in [F.F. Bonsall, J. Duncan, Complete Normed Algebras, Springer–Verlag, Berlin Heidelberg New York 1973].

Given $\|.\|,$ we will define a norm $\|.\|'$ on the field $\mathbb R(X)$ by:
$$\|p/q\|'=\inf_{g}\sup_{h}\|ghp\|/\|ghq\|\tag{•}$$
where the inf and sup range over $\mathbb R[X]\setminus\{0\}.$
We need to check (•) does not depend on the representatives $p,q.$ I will write it as $\inf_g S(g,p,q)$ where $$S(g,p,q):=\sup_h \|ghp\|/\|ghq\|.$$
Note:

*

*$S(g,p,q)\geq S(gk,p,q)$ for any non-zero polynomial $k\neq 0$

*Hence $\inf_g S(g,p,q)=\inf_g S(gk,p,q)$
Also, $S(gq,p',q')=S(gq',p,q)$ if $pq'=p'q.$ Together with property 2, this implies that $\|.\|'$ is well-defined.
For submultiplicativity of $\|.\|'$ consider non-zero $p,q,p',q'.$ Let $\epsilon>0$ be arbitrary.
Pick $g_1,g_2$ such that $S(g_1p',p,q)\leq \|p/q\|'+\epsilon$ and  $S(g_2q,p',q')\leq \|p'/q'\|'+\epsilon$ (making use of property 2). Set $g=g_1g_2.$ Using property 1 we get the same inequalities with $g$ instead of $g_1$ and $g_2.$
The identity $\frac{\|ghpp'\|}{\|ghqq'\|}=\frac{\|ghpp'\|}{\|ghqp'\|}\frac{\|ghqp'\|}{\|ghqq'\|}$ implies $S(g,pp',qq')\leq S(gp',p,q)S(gq,p',q'),$ so $\|pp'/qq'\|'\leq (\|p/q\|'+\epsilon)(\|p'/q'\|'+\epsilon).$ Since $\epsilon$ was arbitrary we get $\|pp'/qq'\|'\leq\|p/q\|'\|p'/q'\|'.$
So $\|.\|'$ makes $\mathbb R(X)$ an infinite-dimensional normed division algebra, which is impossible by the Gelfand-Mazur theorem.
A: I just found this paper: Gelfand–Mazur Theorems in normed algebras: A survey, which contains a "Kaplansky's Theorem":

A real associative algebra with a submultiplicative norm, in which $\lVert fg_n\rVert/\lVert g_n\rVert\to0$ for some sequence $g_n\neq0$ only when $f=0$, is necessarily $\mathbb R$ or $\mathbb C$ or $\mathbb H$.

To apply this, let $L$ and $U$ be lower and upper bounds as in the OP. Clearly $\lVert f\rVert'=U\lVert f\rVert$ is a submultiplicative norm, and
$$\frac{\lVert fg_n\rVert'}{\lVert g_n\rVert'}=\frac{\lVert fg_n\rVert}{\lVert g_n\rVert}\geq L\lVert f\rVert=\frac{L}{U}\lVert f\rVert'$$
cannot approach $0$ unless $f=0$. Therefore $\mathbb R[x]=\mathbb R$ or $\mathbb C$ or $\mathbb H$, a contradiction.
Thus there is no almost multiplicative norm on $\mathbb R[x]$.
