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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.

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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.

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    $\begingroup$ Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$. $\endgroup$ – Kavi Rama Murthy Mar 18 at 8:41
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    $\begingroup$ @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly. $\endgroup$ – Nate Eldredge Mar 19 at 0:36
  • $\begingroup$ What if the polynomial converges to something other than $0$ though? $\endgroup$ – Jam Mar 23 at 16:07
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    $\begingroup$ @Jam You can make it work if $f$ is any function that can be approximated by polynomials. Indeed, if $g_n$ is a sequence of polynomials converging to $f$, then we can assume $g_n(0)$ converges to $f(0)$ since otherwise we're done. But then $p_n+g_n$ still converges to $f$ while $p_n(0)+g_n(0)$ has two subsequences which converge to $f(0)$ and $f(0)+1$ respectively. $\endgroup$ – Ethan MacBrough Mar 23 at 20:47

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