# $\epsilon>0$ there is a polynomial $p$ such that $|f(x)-e^{-x}p|<\epsilon\forall x\in[0,\infty)$

Could any one tell me how to solve this one?

Given $f\in C[0,\infty)$ such that $f(x)\to 0$ as $x\to\infty$ we need to show that for any $\epsilon>0$ there is a polynomial $p$ such that $|f(x)-e^{-x}p(x)|<\epsilon \qquad \forall~ x\in[0,\infty)$

I just know the statement of Weierstrass Polynomial Approximation Theorem and that seems very far from the given problem, but somehow I feel I need to apply the theorem.

• There is one rather complicated solution here. – Milind May 9 '13 at 17:37
• :-o :-o :-o so big solution :-o – Marso May 9 '13 at 17:41
• @milind I upvoted your comment, but I don't agree with the term complicated. The idea is that we can define $f(+\infty):=0$ and we transform continuously $[0,+\infty]$ into $[0,1]$ in order to get a compact interval. Then we approximate. A further approximation (which needs technical verifications) gives what we want. So the given solution is actually conceptually simple, and I would be surprised if a simpler solution would exist. – Davide Giraudo May 10 '13 at 9:47
• @Davide Giraudo Hmm, thank you for the explanation. I confess that I did not understand the basic motivation behind the solution. Does the term "compact" here have a technical meaning? – Milind May 10 '13 at 9:52
• @DavideGiraudo, superb explanation, Such a big proof needs this type of explanations. thanks .. – ROBINSON May 13 '13 at 11:43

$$\vert f(x)-e^{-x}p(x)\vert=e^{-x}\vert g(x)-p(x)\vert,$$ where $\lim_{x\to\infty}e^{-x}g(x)=0$. Your question is about the density of polynomials in the weighted space (weight function is $e^{-x}$) on the half-line. More info can be found Stone-Weierstrass theorem (locally compact version). The literature of weighted approximation in $L_p$ spaces is huge.