Suppose $f:[-1,1] \to \mathbb{R}$ is a continuous function such that $f(0) = f'(0) = 0$. I wish to show that for every $\epsilon > 0$ there exists a polynomial such that
$$\sup_{x\in[-1,1]} |f(x) – x^2p(x)| < \epsilon $$
Letting $\epsilon > 0$, I've immediately used Stone-Weierstrass to show that there is a polynomial $p(x)$ such that $$\sup_{x\in[-1,1]} |f(x) – p(x)| < \epsilon$$
and since $p(x) = \sum_{j=0}^d a_jx^j$, we see that $f(0) = 0$ implies that $a_0 = 0$, so our polynomial is at least linear and non-constant. My idea from here is that $f'(0) = 0$ should imply that the next coefficient $a_1 = 0$ so that our polynomial is at least quadratic, and hence we could factor out an $x^2$, arriving at our claim. How do I prove this last claim?
Best Answer
Your choice of $p$ does not work.
Let $g(x)=\frac {f(x)} {x}$ for $x \neq 0$ and $g(0)=0$. Then $g$ is continuous, so there exists a polynomial $q$ such that $|g(x)-q(x)|<\epsilon$ for all $x$. Note that $ |q(0)| <\epsilon$. If $p_1(x)=q(x)-q(0)$ then $|\frac {f(x)} x-p_1(x)| <2\epsilon$ for all $x >0$. This gives $|f(x)-xp_1(x)| <2\epsilon$. Now note that $p_1(0)=0$ to finish.