How rapidly can a polynomial grow in a proximity of the real segment comparing to the values on the segment? Let $P_n$ be a polynomial of degree $n$ with complex coefficients. Does for any $l>0$ and small $\varepsilon>0$ there exist $C=C(l,\varepsilon)>0$ and $q=q(l,\varepsilon)>1$ s.t.  in the rectangular $D=[-l,l]\times[-\varepsilon,\varepsilon]$ an estimate 
$$
\max_{z\in D} |P_n(z)|
\le Cq^n \max_{x\in[-l,l]}|P_n(x)|
$$
holds and $q(l,\varepsilon)\to1$ as $\varepsilon\to0^+$?
An example of $P_n(z)=z^n$ with $q=(1+\varepsilon^2/l^2)^{1/2}$ shows that $q^n$ cannot be replaced with something growing more slowly.
 A: Let $M=\max_{[-l,l]}|P_n|$ 
The function $u(z)=\log|P_n(z)|$ is subharmonic in $\mathbb C$, and $u(z)=n\log|z|+O(1)$ as $z\to\infty$. 
Since we understand its behavior on $[-l,l]$ and at $\infty$, it makes
sense to apply the three-constant theorem (subharmonic version of the Hadamard three-circle theorem) to $u$.
The map $\phi(z)=(l/2)(z+z^{-1}) $ sends the complement of the unit disk onto the complement of $[-l,l]$. Let $v=u\circ \phi$. The function $v$ is subharmonic in $\{z:|z|>1\}$, satisfies $v\le \log M$ on  the unit circle, and is $n\log |z|+O(1)$ at $\infty$. Since $\max_{|z|=r}v(z)$ is a convex function of $\log r$, it follows that 
$$v(z)\le n\log |z|+\log M \tag{1}$$
Returning to $u$ and then to $P_n$, we conclude that
$$
|P_n(z)|\le M  |\phi^{-1}(z)|^n,\quad z\in\mathbb C\setminus [-l,l] 
\tag{2}$$
Thus, the inequality in your question holds with $C=1$ and 
$$
q = \max_{D} |\phi^{-1}(z)|
\tag{3}$$
which is roughly $1+\sqrt{\epsilon/l}$.  
An aside:  Chebyshev's  inequality (Borwein-Erdelyi, page 235)  states that for every $x\in \mathbb R\setminus [-1,1]$
$$|P_n(x)|\le |T_n(x)|\max_{[-1,1]}|P_n| \tag{4}$$
which is sharp for every such $x$, with equality attained by $P_n=T_n$, the Chebyshev polynomial of 1st kind. The comparison inequality (4) does not extend to complex plane: a complex number $x$ could happen to be close to a zero of $T_n$, in which case (4) fails.   
