Suppose $f\in \mathcal{C}([1,\infty))$ and $\underset{x\rightarrow +\infty}{\lim} f(x)=a$. Show that $f$ can be uniformly approximated on $[1,\infty)$ by functions of the form $g(x)=p(1/x)$, where $p$ is a polynomial.

So I know I need to use the Weierstrass theorem, but I do not know if my approach is right. First, I know that Weierstrass is applied only on intervals of the form $[a,b]$. Would I be able to use it on the interval $[a, a+n]$ and show I can approximate $f$ for any $n\in \mathbb{N}$?

I am also thinking that the end goal here is for my polynomial, $p_n$, to equal $a$ at $x=0$? That way $\underset{x\rightarrow +\infty}{\lim}p_n(1/x)=a$.

If anyone had any hints for the problem and/or can point out if my thinking about this is on the right track, then I'd greatly appreciate it.

  • $\begingroup$ Think about $q\colon[0,1]\to\mathbb{C}$, $$q(t)=\begin{cases}f(1/t)&t>0\\a&t=0\end{cases}$$ $\endgroup$ – user10354138 Nov 14 '18 at 4:10

Hint: Since $f\in \mathcal{C}([1,\infty))$ we have that $f\circ 1/x\in \mathcal{C}((0,1])$. Now, since $\lim_{x\to\infty}f(x)=a$, $\lim_{x\to 0^+}f\circ 1/x=a$. What does this then tell us about the function $$g:[0,1]\to\mathbb{R},\qquad g(x)=\begin{cases} f\circ 1/x & x\in (0,1] \\ a & x=0 \end{cases}?$$


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