How to prove the non-existence of a polynomial series that uniformly converges to this function? I was asked to prove the following. 

There exists a function $p(x)\in C(a,b)$ $(-\infty<a<b<+\infty)$, such that there does not exists a polynomial  series which uniformly converges to $p(x)$ on $(a,b)$. 

I considered $p(x)=e^{\frac1x}$ on $(0,1)$. If there should exist $p_n(x)\rightrightarrows p(x)$ on $(0,1)$, then let $x_n=\frac1n$, I intended to show that $|p_n(1/n)-e^n|\to\infty$ as $n\to\infty$. But I didn't know how to tackle it. Would LHR help? 
 A: Your example certainly suffices, but I don't think you'll have any luck working with the expression $|p_n(\frac{1}n)-e^n|$ since, for appropriate choice of $p_n$, we can make that $0$ for all $n$ (even if it approximates every other point terribly).
The important thing to note is that $p(x)$ is unbounded, whereas all polynomials are bounded. Therefore, for any $n$, we have:
$$\lim_{x\rightarrow 0}|p_n(x)-p(x)|=\infty$$
which contradicts uniform convergence.
A: Let $p_n(x)$ be a sequence of polynomials which converges uniformly to some continuous function $p(x)$ over interval $(a,b)$.
Since the convergence of $p_n(x)$ is uniform, the convergence of $p_n(x)$ over $(a,b)$ is uniformly Cauchy. By defintion, this means
$$\forall \epsilon > 0
\bigg\{ \exists N > 0\bigg[
\forall n, m \ge N, \forall x \in (a,b), \big(\;|p_n(x) - p_m(x)| \le \epsilon\;\big)
\bigg]\bigg\}
$$
Since polynomials are continuous function,
$$|p_n(x) - p_m(x)| \le \epsilon \;\text{ for }\; x \in (a,b)
\implies
|p_n(x) - p_m(x)| \le \epsilon \;\text{ for }\; x \in [a,b]
$$
The means the functions $p_n(x)$ is uniformly Cauchy on $[a,b]$ too.
This implies $p_n(x)$ uniformly converges to some continuous function $\tilde{p}(x)$ over $[a,b]$. 
It is clear $p(x)$ coincides with $\tilde{p}(x)$ on $(a,b)$, As a result, following two limits exists and finite.
$$\begin{align}
p(a^{+}) \stackrel{def}{=} \lim_{x\to a^{+}} p(x) &= \lim_{x\to a^{+}}\tilde{p}(x) = \tilde{p}(a)\\
p(b^{-}) \stackrel{def}{=} \lim_{x\to b^{-}} p(x) &= \lim_{x\to b^{-}}\tilde{p}(x) = \tilde{p}(b)
\end{align}
$$
This also forces $p(x)$ to be bounded on $(a,b)$.

Necessary conditions for $p(x) \in C(a,b)$ to be the uniform limit of a sequence of polynomials
  
  
*
  
*the two limits $p(a^{-})$ and $p(b^{+})$ exist and finite.
  
*$p(x)$ is bounded over $(a,b)$.
  

Now take any unbounded continuous function on $(a,b)$, e.g. $f(x) = \frac{1}{x-a}$, $f(x)$ cannot be the uniform limit of any sequences of polynomials.
