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!

  • $\begingroup$ The function $g$ can be extended with continuity at $y=0$. $\endgroup$ – Rigel Nov 11 '17 at 15:52
  • $\begingroup$ This is a trivial question with the given hint. Define $g(0)$ as $0$, apply the Weierstrass approximation Theorem to $g(x)$, evaluate the polynomial approximations at $x=e^{-t}$. $\endgroup$ – Jack D'Aurizio Nov 11 '17 at 15:53
  • $\begingroup$ thank you for your help I tried to solve below by referring to your advice $\endgroup$ – fivestar Nov 11 '17 at 16:40

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?


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