3
$\begingroup$

I am having trouble showing the following.

Let $f$ be a continuously differentiable function on the closed interval $[0,1]$. Prove that for every $\epsilon >0$ there exists a polynomial $P$ such that \sup_{0 \le x\le 1} |f(x)-P(x)|+\sup_{0 \le x \le 1} |f'(x)-P'(x)| \le \epsilon.

Any hint or suggestion?

  • 0
    @D.Thomine: I don't think it does, so I have changed the tag.2012-03-21

1 Answers 1

4

Hint: You know Weierstrass's approximation theorem, right? Since f' is continuous by assumption, there is some Polynomial $Q$ such that \sup_{0 \le x \le 1} |f'(x) - Q(x)| < \frac{\varepsilon}2 Now integrate ...

HTH, AB,

  • 0
    HTH = "hope this helps", AB = "allzeit bereit" (german Scout motto)2012-03-21