# Does convergence of polynomials imply that of its coefficients?

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.

$$p_n(x)=\left\{ \begin{array}{@{}ll@{}} (1-x)^n, & \text{if}\ n\equiv 0 \mod 2 \\ x^n, & \text{if}\ n\equiv 1 \mod 2 \end{array}\right.$$
Then $$p_n(x)$$ converges to $$0$$ but the constant term is oscillating.
• 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 $x$. – Kavi Rama Murthy Mar 18 at 8:41
• @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly. – Nate Eldredge Mar 19 at 0:36
• What if the polynomial converges to something other than $0$ though? – Jam Mar 23 at 16:07
• @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. – Ethan MacBrough Mar 23 at 20:47