Sequences of polynomial functions converging uniformly on $[a,b]$ to a continuous function not a polynomial

Question

What is (are) the necessary and sufficient condition(s), if any, for a sequence of polynomial functions to converge uniformly on a given (finite) closed interval $[a,b]$ to a continuous function not a polynomial?

In particular, how to show, for example, that the sequence of functions $P_n \colon [a,b] \to \mathbb{R}$ defined for all $t\in[a,b]$ as $$P_n(t) \colon= \sum_{k=0}^n \frac{t^k}{k!}$$ converges uniformly on $[a,b]$ to a function which is not a polynomial?

Answer 1

For any $t\in[a,b]$, the given series is uniformly convergent to $e^t$, which is not a polynomial.

Answer 2

FACT 1: A sequence of polynomials $\{p_{n}\}$ converges uniformly on $[a,b]$ to another polynomial $p$ if the degrees of the polynomials $p_{n}$ s are uniformly bounded and the coefficients are Cauchy sequences.

Proof: Let $\{p_{n}\}$ be a sequence of polynomials whose degrees are uniformly bounded by $D$. Let us write $p_{n}(x)=\sum_{k=0}^{D}a_{n}(k)x^{k}$. Each $\{a_{n}(k)\}_{n}$ is a Cauchy sequence. Therefore there exist real numbers $a(k)\in\mathbb{R}$ such that $a_{n}(k)\to a(k)$ as $n\to\infty$ for all $0\leq k\leq D$. Define a polynomial $p(x)=\sum_{k=0}^{D}a(k)x^{D}$. Notice that for all $x\in[a,b]$

\begin{eqnarray} |p_{n}(x)-p(x)|=\left|\sum_{k=0}^{D}(a_{n}(k)-a_{k})x^{k}\right|\leq C\sum_{k=0}^{D}|a_{n}(k)-a(k)|, \end{eqnarray} where $C=\max_{0\leq k\leq D,x\in[a,b]}|x^{k}|$. Therefore $\|p_{n}-p\|_{\infty}=\sup_{x\in[a,b]}|p_{n}(x)-p(x)|\leq C\sum_{k=0}^{D}|a_{n}(k)-a(k)|\to 0$ as $n\to\infty$, i.e., $p_{n}\to p$ uniformly as $n\to\infty$.

However, I think the converse is not true. For example, take $p_{n}(x)=x+2x^{2}+x^{n-1}-x^{n}$ and $p(x)=x+2x^{2}$. Then $\sup_{x\in[0,1]}|p_{n}(x)-p(x)|=\sup_{x\in[0,1]}|x^{n-1}-x^{n}|=\frac{1}{n}(1-1/n)^{n-1}\to 0$ as $n\to\infty$, i.e., $p_{n}\to p$ uniformly on $[0,1]$. But degrees of $\{p_{n}\}$ are not uniformly bounded.

FACT 2: But if we assume that $\{p_{n}\}$ is a sequence of polynomials of bounded degree and $p_{n}\to p$ uniformly on $[a,b]$, then it can be proved that the coefficients of $p_{n}$ converge to the corresponding coefficients of $p$.

To answer your question: Consider $C^{1}([a,b])$, which is the set of continuously differentiable functions on $[a,b]$. Define the metric $\|\cdot\|_{1}$ on $C^{1}([a,b])$ as $$\|f-g\|_{1}=\sup_{x\in[a,b]}|f(x)-g(x)|+\sup_{x\in[a,b]}|f'(x)-g'(x)|.$$

It is a well known fact that $\left(C^{1}([a,b]),\|\cdot\|_{1}\right)$ is a complete metric space. In our case $P_{n}(x)=\sum_{k=0}^{n}\frac{x^{k}}{k!}\in C^{1}([a,b])$. It is easy to see that $\{P_{n}\}$ is a Cauchy sequence in $C^{1}([a,b])$ under $\|\cdot\|_{\infty}$ norm. Also notice that $P_{n}'=P_{n-1}$. Therefore $\{P_{n}\}$ is also a Cauchy sequence in $C^{1}([a,b])$ under $\|\cdot\|_{1}$ norm. So there exists $f\in C^{1}([a,b])$ such that $P_{n}\to f$ and $P_{n}'\to f'$ uniformly on $[a,b]$. But $P_{n}'=P_{n-1}$, therefore $f'=f$ on $[a,b]$. There is no polynomial function $f$ on $[a,b]$ such that $f'=f$ (because if $f$ is a polynomial the $deg(f')<deg(f)$).