Most general reverse Hölder inequality for polynomials Theorem. Let $m$ be an integer and $P_m$ the vector space of degree $m$ polynomials in one real variable. There is a constant $C$ such that, for all $a<b$ and $p \in P_m$,
$$\|p\|_{L^\infty(a,b)} \leq {C \over b-a} \|p\|_{L^1(a,b)}$$
Proof. Since $P_m$ is finite-dimensional, there is a norm equivalence $\|\cdot\|_{L^\infty(0,1)} \leq C \|\cdot\|_{L^1(0,1)}$. The corresponding inequality follows by change of variable $x = (b-a)y+a$. □
I'd like to generalize this as much as possible, but I'm not sure how much is possible. One possible generalization is to all measurable sets $E$:
$$\|p\|_{L^\infty(E)} \leq {C \over |E|} \|p\|_{L^1(E)}?$$
The given Theorem is the special case where $E$ is a single interval. Maybe it's necessary to impose that $E$ is in the interval, to avoid some trivialities.
I think the following generalization might be false though. You could allow $\mu$ to vary over probability measures and ask for:
$$\|p\|_{L^\infty(d\mu)} \leq C \|p\|_{L^1(d\mu)}?$$
I tried to look in this direction, thinking of the Banach-Alaoglu theorem.
I'm marking this as "Complex Variables" because reverse Hölder inequalities appear in complex analysis textbooks.
 A: Your second version is not true for arbitrary $E \subset [0,1]$.  Consider $m=1$, $f(x) = x$ and
$E = [0,\epsilon - \epsilon^2] \cup [1-\epsilon^2, 1]$ where $0 < \epsilon < 1$.  We have $|E|=\epsilon$, $\|f\|_{L^\infty(E)} = 1$ but
$\|f\|_{L^1(E)} =3 \epsilon^2/2 - \epsilon^3$,
so $\|f\|_{L^1(E)}/|E| \to 0$ as $\epsilon \to 0+$.
A: Although Robert Israel's answer is completely accurate, I was able to find the generalization I was looking for. In case someone is looking for this exact thing in the future, I'm writing it up here.
Theorem. Let $m$ be an integer and $P_m$ the space of polynomials of degree $m$ in one real variable. Let $\kappa > 0$. There is a constant $C$ such that the following holds. If $a<b$ and $d\mu = g(x) dx$ is a Radon probability measure with density function $g(x)$ on $[a,b]$, such that $\|g\|_{L^{\infty}(a,b)} \leq {\kappa \over b-a}$, and $p \in P_m$, then
\begin{align}
\|p\|_{L^{\infty}(a,b)} \leq C \|p\|_{L^1(d\mu)}.
\end{align}
Proof. We begin with the case $a=0$ and $b=1$. By Alaoglu's theorem, the set $Q = \{ g \in L^{\infty}(a,b) \; : \; \|g\|_{L^{\infty}(a,b)} \leq \kappa \text{ and } g \geq 0 \text{ and } \int_a^b g \,dx = 1 \}$ is compact in the weak-* topology of $L^{\infty}(a,b)$ as the dual of $L^1(a,b)$. Put
\begin{align}
\phi(f,\mu) & = \int |f| \, d\mu.
\end{align}
Denote by $S \subset P_m$ the unit sphere defined by $\|p\|_{L^{\infty}(0,1)} = 1$. On the set $S\times Q$, we impose the topology $(\text{uniform}) \times (\text{weak}-*)$.
We show that $\phi$ is continuous on $S \times Q$ in this topology. To that end, assume that $\mu_k \to \mu_{\infty}$ in the weak-* topology, and $p_k \to p_{\infty}$ uniformly. Then,
\begin{align}
|\phi(p_k,\mu_k) - \phi(p_{\infty},\mu_{\infty})| & \leq
|\phi(p_\infty,\mu_k) - \phi(p_\infty,\mu_{\infty})|
+|\phi(p_\infty,\mu_k) - \phi(p_k,\mu_k)|
\end{align}
On the right-hand-side, the first term $\to 0$ by weak-* convergence of $\mu_k$, and the second term is bounded by
\begin{align}
|\phi(p_\infty,\mu_k) - \phi(p_k,\mu_k)|
& = \left|
\int |p_{\infty}| - |p_k| \, d\mu_k
\right|
\leq
\int |p_{\infty} - p_k| \, d\mu_k
\leq \|p_{\infty} - p_k\|_{L^{\infty}(0,1)}, 
\end{align}
which also $\to 0$ as $k \to \infty$. Thus, $\phi$ is continuous on the compact set $S\times Q$ and hence it attains a minimum, say at $(p_0,\mu_0)$. If $\phi(p_0,\mu_0) = 0$ then $\|p_0\|_{L^1(d\mu_0)} = \int |p_0(x)| g_0(x) \, dx = 0$, so that $p_0(x)g_0(x) = 0$ a.e. Because $d\mu_0 = g_0(x) \, dx$ is a probability density function, it must be that $\int g_0(x) \, dx = 1$, and hence $g_0(x) \neq 0$ on a set of positive measure, thus $p_0(x)=0$ on a set of positive measure. In particular, $p_0(x) = 0$ for infinitely many values of $x$. Since $p_0 \in P_m$, it must be that $p_0=0$, contradicting $\|p_0\|_{L^{\infty}(0,1)} = 1$. Thus, it must be that $\phi(p,\mu) \geq \epsilon>0$ for all $(p,\mu) \in S \times Q$. By homogeneity in $p$, we find that
\begin{align}
\|p\|_{L^{\infty}(0,1)} \leq \epsilon^{-1}\phi(p,\mu) = \epsilon^{-1} \|p\|_{L^1(d\mu)}.
\end{align}
so that $C = \epsilon^{-1}$.
For generic $a<b$ and probability measure $d\nu = h(y) \, dy$ on $[a,b]$, we perform the substitution $p(x) = q((x-a)/(b-a)) = q(y)$ and $(b-a)h(y)dy = g(x)dx$ and use the scalings
\begin{align}
\|q\|_{L^{\infty}(a,b)} = \|p\|_{L^{\infty}(0,1)} \leq C \|p\|_{L^1(d\mu)} = C\|q\|_{L^1(d\nu)}.
\end{align}
□
Maybe the following corollary is more aesthetically pleasing though:
\begin{align}
\|p\|_{L^{\infty}(d\mu)} \leq C \|p\|_{L^1(d\mu)}.
\end{align}
However, I find this version confusing because it obfuscates the interval $[a,b]$, and that interval cannot be dispensed with, essentially because of Robert Israel's example.
