Question: Let $K\subset\mathbb{C}$ be a convex set and $F_K=\{f\in\mathbb{C}[x]:f(z)\subset{}K, \forall{}z\in{}D\}$ where $D=\{z\in\mathbb{C}:|z|<1\}$. Let $\hat{P}(n)$ denote the $n$th coefficient of the polynomial $P$. If $E_n=\{a\in\mathbb{C}:\exists{}P\in{}F_K:\hat{P}(n)=a\}$ show that: $E_0=K$ and $E_1=E_2=...=E_i=..$ for every $1\leq{}i\in\mathbb{N}$.
Attempt at Answer: I am not sure how to approach this, it is obvious that $E_0=K$ since all the constant polynomials $P(x)=k$ will work for every $k\in{}K$.
To show $E_i\subseteq{}E_{i+1}$ I had the following idea: if $P(z)\subset{}K,\forall{}z\in{}D$ then the polynomial $f(z)=P(z)-a_0\subset{}K-a_0,\forall{}z\in{}D$ and $f(0)=0$. Now let $g(z)=zf(z)$. Then $|g(z)|<|f(z)|$ for $z\in{}D$ and thus every point $g(z_0)$ is contained in the line segment connecting $0$ and $f(z_0)$ with $|z_0|=1$ which is contained in $K-a_0$ since it is also convex. Then $zf(z)\subset{}K-a_0 \implies z(P(z)-a_0)+a_0\subset{}K$. So if $Q(z)=z(P(z)-a_0)+a_0$ and $P\in{}F_K, \hat{P}(n)=a$ then $Q\in{}F_K$ and $\hat{Q}(n+1)=a$.
This proves $E_i\subseteq{}E_{i+1}$ so now I need to show $E_{i+1}\subseteq{}E_i$. How can I do this?
We assume $F_K$ has nonconstant polynomials as otherwise nothing to prove (that happens iff $K$ has nonempty interior by the open map property of polynomials and easy considerations for sufficiency - eg if $w$ is an interior point of $K$, $w+\frac{z}{M}$ is nonconstant in $F_K$ for all large enough $M>0$ say).
Then we show $E_1=E_n$ for $n \ge 2$. One way is trivial and doesn't use convexity since obviously by taking constants all $E_k$ contain zero for $k \ge 1$; so if $a \ne 0$ in $E_1$ is realized by some nonconstant $f$, $f(z^n)$ realizes $a$ in $E_n$ and has same image as $f$ on the unit disc.
The other inclusion needs convexity. Let $a \ne0$ in $E_n, n \ge 2$ and $f \in F_K$ realizing it. Then since the rotations of $f$ are in $F_k$, $f(w_kz)$ is there for all $w_k$ roots of unity of order $n$, and since $K$ is convex, the arithmetic mean of those $n$ polynomials is in $F_K$. But said arithmetic mean is a function of $z^n$ as only the coefficients with index that divides by $n$ remain unchanged and all the others become zero. Substituting $z^n$ with $z$ gives that $a$ is in $E_1$. Done!