**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.



Most general reverse Hölder inequality for polynomials
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.


