4

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?

troutman314
  • 81
  • Your choice of $p$ does not work. – Kavi Rama Murthy Jul 28 '21 at 23:14
  • There is no reason for this last claim to be true. For any $\varepsilon>0$, you can have some tiny noise in the linear and constant terms. I see two ways of proceeding: either show that you can first remove the linear and the constant term without harming the estimate too much, or show that $f/x^2$ is, in fact, continuous. – tomasz Jul 28 '21 at 23:15
  • Please explicitly state the Stone-Weierstrass Theorem in the most general form you know. The are various versions and it is important to see which can be used. – Paul Frost Jul 28 '21 at 23:16
  • @PaulFrost: I can kind of see a solution using some stronger version of Stone-Weierstrass, but I somehow doubt it'd be any simpler than a solution using the most basic one. – tomasz Jul 28 '21 at 23:17
  • @KaviRamaMurthy: I think it's pretty clear: he says explicitly that $f'(0)=0$ (so in particular, it exists). – tomasz Jul 28 '21 at 23:18
  • 1
    @tomasz $f(x)/x^2$ does in general not have a continuous extension to $[-1.1]$. Take $f(x) = x^2\sin(1/x)$ for $x \ne 0$ and $f(0) = 0$. Either we need additional assumptions on $f$ or another approach. – Paul Frost Jul 29 '21 at 11:36

1 Answers1

5

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.

Kavi Rama Murthy
  • 311,013