Show that there exist a sequence of function $g_n$ that converge uniformly to $f$

Question

Let $f:[1,\infty[$ be a continuous function such that $lim_{x \to > \infty}f(x)=a$. Show that there exist a sequence of function $g_n$ that converge uniformly to $f$ where $g_n(x)=P_n(1/x)$ with $P_n$ a polynomial.

To solve this probleme I think we need to use the approximation of Weierstrass but the fucntion $f$ is not a compact .

I dont know how to proceed and I'm not sure how to use the approximation of Weierstrass theorem.

Answer 1

Let $g(x)= \begin{cases} f(1/x) & 0<x\leq 1 \\ a & x=0 \end{cases} $

$g(x)$ is continuous on $(0,1]$ because $f(x)$ is a continuous function and also $g(x)$ is continuous at $x=0$. It follow that $g$ is continuous on $[0,1]$.

By the Weierstrass Approximation theorem, there exist a polynomial $P_n(x)$ that converge uniformly to $g(x)$ on $[0,1]$.

Since $\forall x \in[0, \infty)$: $$|g(1/x) - P_n(1/x)| =|f(x) - P_n(1/x)| $$

it follow that $\forall \epsilon >0 , \exists N \in \mathbb{N}$ such that $n>N$:

$$|f(x) - P_n(1/x)|<\epsilon .$$

We conclude that $P_n(1/x)$ converge uniformly to $f(x)$ on $[0,\infty)$.

Answer 2

Consider the function $g$ on $[0,1]$ given by $g(x) = f(1/x)$ for $x > 0$, $g(0) = a$, and approximate it by polynomials using Weierstrass.