Behavior of coefficients of sequence of polynomials converging to the zero function

Question

Consider a sequence of polynomial functions $\{f_n(x)\}$ defined on a closed interval $[a,b]$ with $0 < a < b < 1$.

The function $f_n(x)$ is of the form $$ f_n(x) = a_{n1} x + a_{n2} x^2 + ... + a_{nn} x^n. $$

As $n \to \infty $, the sequence $\{f_n(x)\}$ converges uniformly to the zero function over $[a,b]$.

Question: Does it follow that for every $j$, we have $$ \lim_{n \rightarrow \infty} a_{nj} = 0?$$

I note that if the degree of $f_n(x)$ is bounded above by $M$ (for all $n$), then of course the answer is in the affirmative (this question was asked and answered in an earlier posting by someone else). That proof used the fact that the associated vector space is finite dimensional, which is not the case here.

Answer 1

Counterexample: By Weierstrass there are polynomials $p_n(x)$ converging uniformly to $\sqrt x$ on $[0,1].$ Let $d_n$ be the degree of $p_n.$ If the set $\{d_n\}$ were bounded, say by $D,$ then the $p_n$ would converge uniformly to a polynomial of degree $\le D.$ But $\sqrt x$ is not a polynomial, contradiction.

Thus we can choose a subsequence $d_{n_k}$ such that $n_1<n_2 < \cdots $ and $d_{n_1} < d_{n_2}\cdots \to \infty.$ We then have $p_{n_k}(x) \to \sqrt x$ uniformly on $[0,1],$ which implies $p_{n_k}(x^2) \to x$ uniformly on $[0,1].$

So we have $p_{n_k}(x^2) - x$ converging uniformly to $0$ on $[0,1].$ But notice that the expansion of each $p_{n_k}(x^2)-x$ has $-1$ for the coefficient of $x.$ We thus have a counterexample, save for a small detail. The degree of each $p_{n_k}$ is $2d_{n_k}.$ So we form a new sequence $q_n,$ where $q_n=0$ if $n$ is not one of the $d_{n_k},$ while $q_n = p_{n_k}(x^2)-x$ if $n= 2d_{n_k}.$

Answer 2

No, that does not follow.

Approximate the function $-{1\over x}$ in the interval $[a,b]$ by polynomials. You can do this with arbitrary small supremum distance according to the (Stone-)Weierstrass theorem: https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem, so let

$$\lvert p_n(x) + {1 \over x}\rvert < {1 \over n}, \forall x \in [a,b]$$

for some polynomial $p_n(x)$. If necessary, add some $a_{r(n)}x^{r(n)}$ term with high $r$ and small $a_r$ to $p_n(x)$ such that the inequality above still holds and we have

$$\deg p_n(x) < \deg p_{n+1}(x), \forall n=1,2,\ldots$$

Now define your functions $f_n(x)$ as follows:

$$ f_n(x) = \begin{cases} x+x^2p_k(x), & \text{if}\; n=\deg p_k(x)-2 \\ 0, & \text{otherwise} \end{cases} $$

We have $\lvert x+x^2p_k(x)\rvert = x^2\lvert {1\over x}+p_k(x)\rvert < 1\times{1\over k}$, as $0 < x < 1$, so $f_n(x)$ uniformly converges to $0$, but has inifintely many elements where the coefficient of the linear term is $1$.