How to prove the non-existence of a polynomial series that uniformly converges to this function?

Question

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?

Answer 1

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.

Answer 2

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.