Must convergent sequence of polynomials have bounded coefficients? Given $$P_n = a_0^{(n)}+...+a_k^{(n)}x^k$$ where $k$ is fixed independent of $n$, such that $$P_n \to f$$ with respect to the uniform norm. Then must we have that the coefficients are uniformly bounded over $n$? I am sure that this is true, but have been unable to solve it.
 A: Since $P_n(x)$ is continuous and the convergence is uniform we have that $f$ is a continuous function. Since $P_n(x)$ is a polynomial with degree $\leq k$ we have
$$ \sum_{j=0}^{k+1}\binom{k+1}{j}(-1)^j P_n(x+j\varepsilon) = 0 $$
for any $n\in\mathbb{N}, x\in\mathbb{R}$ and $\varepsilon\in\mathbb{R}\setminus\{0\}$, by finite differences. By uniform convergence the same identity has to hold if $P_n$ is replaced by $f$: in particular $f$ behaves like a polynomial over $\varepsilon\mathbb{Z}$ and any translate of $\varepsilon\mathbb{Z}$, for any $\varepsilon>0$. Continuity easily leads to the conclusion that $f$ is a polynomial, hence the sequences $\{a_0^{(n)}\}_{n\geq 1}, \{a_1^{(n)}\}_{n\geq 1}, \ldots \{a_k^{(n)}\}_{n\geq 1}$ are convergent and uniformly bounded.

If we are dealing with uniform convergence over $[0,1]$, we may notice that there is an invertible linear map bringing $v^{(n)}=(a_0^{(n)},\ldots,a_k^{(n)})$ into $w^{(n)}=(b_0^{(n)},\ldots,b_k^{(n)})$ such that
$$P_n(x) = \sum_{j=0}^{k}b_j^{(n)}Q_j(x) $$
where $Q_j(x)$ are the normalized shifted Legendre polynomials, giving an orthonormal base of $L^2(0,1)$. If some sequence among $\{a_0^{(n)}\}_{n\geq 1}, \{a_1^{(n)}\}_{n\geq 1}, \ldots \{a_k^{(n)}\}_{n\geq 1}$ were unbounded, the same would apply to some $\{b_j^{(n)}\}_{n\geq 1}$, leading to an unbounded dot product between $P_n(x)$ and $Q_j(x)$. By uniform convergence this gives an unbounded dot product between $f(x)$ and $Q_j(x)$, contradicting
$$ \left|\int_{0}^{1}f(x)Q_j(x)\,dx\right|\leq \sqrt{\int_{0}^{1}f(x)^2\,dx}\leq \max_{x\in[0,1]}|f(x)|.$$
A: Another answer:
Let $E=C([0,1])$ be the space of continuous functions on $[0,1]$ with the $L^\infty$-norm. Then the subspace $F$ of polynomial functions with degree at most $k$ is finite-dimensional. Since finite-dimensional subspaces are closed, $F$ is closed in $E$. So, the limit $f$ is in $E$ : it is a polynomial function with degree at most $k$. Finally, since in finite dimensional vector spaces, all the norms are equivalent, the convergence in $F$ in $L^\infty$ norm implies the convergence in the norm $|p|=\max(|a_0|,\dots,|a_n|)$ (for instance). Thus, all the coefficients converge, and are in particular bounded.
(And after reading the question linked by Robert Z., I realize my answer was already there.)
