I'm trying to show that for every real polynomial $p$ of degree smaller or equal to some $d \in \mathbb{N}$, and a set $S\subset \mathbb{R}$ with more than $d+1$ points, there is a constant $C_{d,S} = C(d,S)$, which depends only on $d$ and $S$, such that:
$$\forall p \in \mathcal{P}_d : \max(|a_i(p)|:0\le i \le d) \le C(d,S)\sup{({|p(x)|:x\in S})}$$
where $p = \sum_{i=0}^d a_i(p)x^i$, and $\mathcal{P}_d$ is the space of such polynomials. The inverse inequality seems to come naturally by triangle inequality, if this set $S$ was bounded, but this one doesn't look that easy. Any advice will be very welcome. Thanks in advance :).
It suffices that the set $S$ has least $d+1$ distinct points $x_0, x_1 \ldots, x_d$.
Every polynomial $p$ of degree at most $d$ is equal to the interpolation polynomial for the data $(x_k, p(x_k))$, $k=0, \ldots, d$: $$ p(x) = \sum_{j=0}^d p(x_j) L_{d, j}(x) $$ where $$ L_{d, j}(x) = \prod_{\substack{k=0 \\ k \ne j}}^d \frac{x-x_k}{x_j - x_j} $$ are the Lagrange polynomials for $x_0, \ldots x_d$. Then $$ |a_i(p)| = \left|\frac{1}{i!} p^{(i)}(0)\right| = \frac{1}{i!}\left|\sum_{j=0}^d p(x_j) L_{d, j}^{(i)}(0)\right| \\ \le C_i(d, S) \cdot \sup \{ p(x) : x \in S\} $$ where $$ C_i(d, S) = \frac{1}{i! }\sum_{j=0}^d \left|L_{d, j}^{(i)}(0)\right| \, . $$ depends only on the points $x_0, \ldots, x_d$.
$C(d, S) = \max \{ C_i(d,S) : 0 \le i \le d \}$ then does the job.