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:
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.
