Restriction of polynomials of order $m$ to $(t_1, \ldots, t_n)$ is closed? Let $\{p_k \}_k$ denote a sequence of polynomials of order $m$ and let $\{t_1, \ldots, t_n\}$ denote $n$ distinct points in $[0,1]$ with $n \geq m$. I know that 
$$ \lim_k p_k(t_i) \quad \text{exists}, \quad i = 1, \ldots, n $$
Can I conclude that the limit is a polynomial? It would seem to me that this is the case since the vector space of polynomials of order $m$ is finite-dimensional and hence closed.  If this is true, could someone point me to a relevant formal result? 
 A: If $n < m+1$ then the claim is not true: One can construct a non-zero polynomial $p$ of degree $m$ that is zero at all points $t_i$. Then set $p_k:=k\cdot p$.
Let me suppose $m+1=n$. 
Let me denote the vector space of polynomials of degree at most $m$ by $V$ supplied with some norm. Then $dim(V)=m+1$.
The evaluation functionals $\phi_i(p):=p(t_i)$ are linearly independent (use Lagrange polynomials). Since  there are $m+1$ such functionals, and the dimension of the dual space $V^*$ of $V$ is equal to the dimension of $V$, these functionals form a basis of $V^*$.
Using Lagrange polynomials again, there is exactly one polynomial $p$ of degree at most $m$ such that $\phi_i(p) = \lim_k \phi_i(p_k)$ for all $i$.
Let now $\phi\in V^*$ be given. Then $\phi$ is a finite linear combination of these $(\phi_i)$, and it follows $\phi(p_k) \to \phi(p)$.
This implies weak convergence of $(p_k)$ to  $p$.
Since the underlying vector space is finite-dimensional, we have strong convergence of $p_k$ to $p$.

Another possible proof is the following one: The map $p\mapsto \sum_i |\phi_i(p)|$ is a norm on $V$ (this needs $m+1\le n$). Then $(p_k)$ is a Cauchy sequence in this norm.
But $V$ is finite-dimensional, hence complete, and $p_k$ is converging.
