Denote by $D(r)$ the disc at the origin of radius $r>1$. Denote by $P\_m$ the set of polynomials of degree $m$. Since $P\_m$ is finite dimensional, there is a constant $C(m,r)$ such that
$$
\|p\|\_{L^{\infty}(D(r))}
\leq
C(m,r)
\|p\|\_{L^1(-1,1)}
$$
for all $p \in P\_m$.


I'd like some estimates for $C(m,r)$, especially the growth in $m$?



The constant of the reverse Hölder inequality for polynomials
I presume that $D(r)$ refers to a disk in the complex plane, in which case the maximum value of a polynomial (like any analytic function) is achieved on the boundary of $D(r)$. If $\|p\|\_{L^\infty(D(r))} \le C \|p\|\_{L^1(-1,1)}$, it means that the ball $\|p\|\_{L^1(-1,1)} \le 1/C$ fits entirely into the ball $\|p\|\_{L^\infty(D(r))}\le 1$ and by convexity it is enough to check that condition for the extreme points of the $L^1(-1,1)$-ball. So bounding $C$ from above is the same as bounding $\|p\|\_{L^1(-1,1)}$ from below with $\|p\|\_{L^\infty(D(r))} = 1$.


By a result of



> 
> *Garkavi, A. L.*, The unit sphere of the space of polynomials with integral metric, [Mat. Zametki 1, 299-304 (1967)](http://mi.mathnet.ru/mzm9417). [ZBL0156.36603](https://zbmath.org/?q=an:0156.36603), [MR209815](https://mathscinet.ams.org/mathscinet/article?mr=209815).
> 
> 
> 


the extreme points of the $L^1(-1,1)$-ball for polynomials of degree $m$ are those that have exactly $m$ roots in the interval $[-1,1]$ counting multiplicity. We can represent such a polynomial with $\|p\|\_{L^\infty(D(r))} = 1$ as
$$
 p(x) = \frac{1}{N} (x-a\_1) \cdots (x-a\_m) ,
$$
where the normalization factor $N=(z-a\_1)\cdots(z-a\_m)$ for some complex $|z|=1$ and all the roots $a\_i \in [-1,1]$, while
$$
 \|p\|\_{L^1(-1,1)} = \frac{2}{N} \int\_{-1}^1 \prod\_{i=1}^m |x-a\_i| \frac{dx}{2} \ge \frac{1}{C} .
$$
So to bound $1/C$ from below, it is enough to bound each factor from below.


For $N$, it is easy to see that $1/N \ge 1/(r+1)^m$, realized by $p(x) = (x+1)^n$ for instance.


For the integral, using the convexity of $\exp(x)$, we get
$$\begin{aligned}
 \int\_{-1}^1 \prod\_{i=1}^m |x-a\_i| \frac{dx}{2}
 &\ge \exp\left(\int\_{-1}^1 \sum\_{i=1}^m \ln|x-a\_i| \frac{dx}{2}\right) \\
 &= \exp\left(\frac{1}{2} \sum\_{i=1}^m \left[(1-a\_i)\ln|(1-a\_i)/e|+(1+a\_i)\ln|(1+a\_i)/e|\right]\right) \\
 &\ge \exp(-m) ,
\end{aligned}$$
where the last inequality is saturated when all $a\_i=0$.


Putting together the pieces, we find that
$$
 C \le 2 e^m (r+1)^m .
$$
This ~~improves the bound from~~ is an alternative bound to [the other answer](https://mathoverflow.net/a/450139). It gets rid of the the $(m+1)$ factor, but is is generally worse because the base exponent $e(r+1) > (2r+1)$.


