Let $\alpha\gt 0$ .Using the Weierstrass Theorem, prove that every continuous function $f(x)$ on [0,$\infty$] with $\lim_{x \to \infty} f(x) = 0$ can be uniformly approximated as closely as we like by a function of the form $q(x)=\sum_{n=1}^{N}C_n e^{-n\alpha x}$. Hint consider $g(y)= f(-\log(y)/\alpha)$ on $(0,1]$
Though there is hint, I can't solve this problem anyhow. I want to know how to start to solve this problem. I'm waiting your help. thank you!
I tried to solve by referring to your help.
Let $g(y)=f(-\log(y) /\alpha)$ and $g(0)=0$
then by weierstrass approximation thm. there is some polynomial $p_n(y) =\sum_{k=0}^{n} C_k y^k$ such that $p_n \to g $ ,as $ n \to \infty,$ uniformly.
and let $y=e^{-\alpha x}$ then $f(-\log e^{-\alpha x}/a)=f(x)$ and $q(x)=p_n(e^{-\alpha x})=\sum_{k=0}^{n} C_k e^{-\alpha xk}$
thus $f(x)$ can be unifromly approximated by form of $q(x)$
is there no fault?