I was working in the next exercise but I don't know what to do. The exercise is the next:
Let $f:[0,1]\to\mathbb{R}$ be a continuous function and suposse that for some $\alpha>1$ the inequality $$\left|\dfrac{f(x)}{x^{\alpha}} \right|\leq 1$$holds for all $x\in (0,1]$. Prove that $f$ can be aproximated uniformly by polynomials of the form $a_2x^{2}+\cdots+a_nx^{n}$ with $n\geq 2$.
I really don't know how to proced. My first idea was to think in Stone-Weierstrass theorem because then I can find a sequence of polynomials that converges uniformly to $f$ but the problem is the degree. How can I "delete" the first terms of the polynomials and don't lose the convergence? Is it possible? Anothe idea was to think in Taylor series but I'm not sure that $f$ is an infinitely many differentiable function. Any hint? I really aprecciate any help you can provide me. Thanks.
Best Answer
The set of all polynomials $p$ with $p'(0)=0$ is a subalgebra of $C(0,1]$ which contains constants and separates points. Hence, given any continuous function $f$ and $\epsilon >0$ we can find a polynomial $p$ such that $p'(0)=0$ and $|f(x)-p(x)| <\epsilon $ for all $x$. Note that $|p(0)| <\epsilon$ if $f(0)=0$. Let $q(x)=p(x)-p(0)$. Then $q(0)=q'(0)=0$ and $|f(x)-q(x)| <2\epsilon$ for all $x$.
I did not use the full force of the hypothesis. I only need $f(0)=0$.
A much simpler proof: There is a polynomial $p$ such that $f(\sqrt x)-p(x)| <\epsilon $ for all $x$. If $q(x)=p(x)-p(0)$ then $|f(\sqrt x)-q(x)| <2\epsilon $ for all $x$. Just replace $x$ by $x^{2}$ to finish the proof.