Extending Uniform Convergence of Polynomials From $\left[0,1\right]$ to $\mathbb{\overline{\mathbb{D}}}$ This is a question that has been bothering me in my research to no end:
Let $\varphi$
  and the sequence $\left\{ \varphi_{n}\right\}$ 
  be complex-valued polynomial functions defined on the closed unit interval $\left[0,1\right]\subseteq\mathbb{R}$
  such that: $$\lim_{n\rightarrow\infty}\underset{x\in\left[0,1\right]}{\sup}\left|\varphi\left(x\right)-\varphi_{n}\left(x\right)\right|=0$$
 i.e., the polynomials $\varphi_{n}$
  converge uniformly to a polynomial $\varphi$
  on $\left[0,1\right]$
 . Since polynomials always extend to entire functions, it seems almost obvious that the given convergence on $\left[0,1\right]$
  would imply uniform convergence on the closed unit disk $\overline{\mathbb{D}}$
 :$$\lim_{n\rightarrow\infty}\underset{\left|z\right|\leq1}{\sup}\left|\varphi\left(z\right)-\varphi_{n}\left(z\right)\right|=0$$
 However, I am not entirely sure if this is true, nor how to prove it. 
I have showed, however—by assuming otherwise and exploiting Montel's Theorem using the assumption $\left|\varphi_{n}\left(z\right)-\varphi\left(z\right)\right|>\epsilon$
  for all $z$
  in an open set $U\subseteq\mathbb{D}\backslash\left[0,1\right]$—that: $$\lim_{n\rightarrow\infty}\left|\varphi\left(z\right)-\varphi_{n}\left(z\right)\right|=0$$
 pointwise for all $z\in\overline{\mathbb{D}}$.
Clarification on this matter would be greatly appreciated.
 A: Counterexample: For $n=1,2,\dots$ we can choose a polynomial $p_n$ such that
$$|p_n(z) - \frac{e^{inz}}{n}| \le \frac{1}{n}\,\,\text {for } z\in \overline {\mathbb D}.$$
(For example, we could take $p_n$ to be a Taylor polynomial of $\dfrac{e^{inz}}{n}$ at $0$ of sufficiently high degree.)  It follows that $p_n\to 0$ uniformly on $[-1,1].$ However, if $z= x+iy$ with $y<0,$ then $|\dfrac{e^{inz}}{n}|= e^{n|y|}/n \to \infty.$ Thus for any $z\in \overline {\mathbb D}$ above the real axis, we will have $|p_n(z)|\to \infty.$
A: There actually is a way to accomplish this extension: use the Poisson Kernel.
Instead of considering a sequence of polynomials $\varphi_{n}$
  in $x$
  on the unit interval, consider a sequence of polynomials $\varphi_{n}$
  in $e^{2\pi it}$
  for $t$
  in the unit interval (such that $\varphi_{n}\left(0\right)=\varphi_{n}\left(1\right)$
  for all $n$
 ). Then, by the Poisson integral formula, we can extend both the $\varphi_{n}$s and their limit, $\varphi$
 , to holomorphic functions $\psi_{n}$
  and $\psi$
 :$$\psi_{n}\left(z\right)=\int_{0}^{1}\varphi_{n}\left(t\right)\textrm{Re}\left(\frac{e^{2\pi it}+z}{e^{2\pi it}-z}\right)dt,\textrm{ }\forall\left|z\right|<1$$
 and:$$\psi\left(z\right)=\int_{0}^{1}\varphi\left(t\right)\textrm{Re}\left(\frac{e^{2\pi it}+z}{e^{2\pi it}-z}\right)dt,\textrm{ }\forall\left|z\right|<1$$
 Since the poisson kernel is bounded on all compact subsets of $\mathbb{D}$
 , and since the $\varphi_{n}$s converge uniformly in $t$
 , $\psi_{n}$
  converges compactly to $\psi$ on the open unit disk. 
Thus, the moral of the story is that you can extend the polynomials' limit to a holomorphic function by way of that function's values the unit circle.
