# A real analysis question

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?

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Hint: You know, that polynomials are dense in $C[0,1]$, right? Now $f'$ is continuous, so there is some polynomial such that your second term is $< \epsilon/2$ ... now integrate. –  martini Mar 21 '12 at 17:08
I'm a bit curious: how does this question relate to measure theory? –  D. Thomine Mar 21 '12 at 17:10
@martini: why don't you write that as an answer? –  Martin Argerami Mar 21 '12 at 17:14
@D.Thomine: I don't think it does, so I have changed the tag. –  Martin Argerami Mar 21 '12 at 17:16

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,

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thanks! I got it! but just curious what is HTH and AB? –  Alex J. Mar 21 '12 at 19:29
HTH = "hope this helps", AB = "allzeit bereit" (german Scout motto) –  martini Mar 21 '12 at 22:12