TITLE: Does convergence of polynomials imply that of its coefficients?
QUESTION [46 upvotes]: Let $\{p_{n}\}$ be a sequence of polynomials and $f$ a continuous function
on $[0,1]$ such that $\int\limits_{0}^{1}|p_{n}(x)-f(x)|dx\to 0$.
Let $c_{n,k}$ be the coefficient of $x^{k}$ in $p_{n}(x)$. Can we conclude
that $\underset{n\rightarrow \infty }{\lim }c_{n,k}$ exists for each $k$?.
What I know so far: if the degrees of $p_{n}^{\prime }s$ are bounded then
this is true. In fact, we can replace $L^{1}$ convergence by convergence in
any norm on $C[0,1]$; to see this we just have to note that for fixed $N$, $%
\sum_{k=0}^{N}c_{i}x^{i}\rightarrow (c_{0},c_{1},...,c_{N})$ is a
linear map on a finite-dimensional subspace and hence it is continuous. My
guess is that the result fails when there is no restriction on the degrees.
But if $p_{n}(z)$ converges uniformly in some disk around $0$ in the complex
plane then the conclusion holds. To construct a counterexample we have to
avoid this situation. Maybe there is a very simple example but I haven't been
to find one. Thank you for investing your time on this.

REPLY [56 votes]: Consider the sequence of polynomials
$$
p_n(x)=\left\{
  \begin{array}{@{}ll@{}}
    (1-x)^n, & \text{if}\ n\equiv 0 \mod 2 \\
    x^n, & \text{if}\ n\equiv 1 \mod 2
  \end{array}\right.
$$
Then $p_n(x)$ converges to $0$ but the constant term is oscillating.