Aproximating function by polynomials of degree at least $2$.

Question

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.

Answer 1

The Bernstein polynomials

Use the Bernstein proof of the Weierstrass theorem for your purpose.

Indeed, let $f \in C[0,1]$ and $f_n(x) = \sum_{k=0}^n \binom nk x^k(1-x)^{n-k}f(\frac kn)$. Then $f_n$ are polynomials. It is well known that $f_n(x) \to f(x)$ uniformly on $[0,1]$, a proof of this can be found here. If you do not know this fact, then you have just been enlightened! $f_n$ is called the $n$th Bernstein polynomial corresponding to $f$, and the attached proof is called Bernstein's proof of the Weierstrass theorem.

The Basic Idea

We will try to obtain our polynomial by removing constants and terms having just $x$ (and not higher powers) from this summation. Now, $f_n$ is a polynomial of degree $n$. Note, however, that if $k \geq 2$ then the term corresponding to $k$ has powers of $x$ only above $x^2$, because we are multiplying by $x^k$ with $k \geq 2$. So we can leave these terms as they are.

The term for $k=0$ is $(1-x)^n f(0)$. However, $|f(x)| \leq |x|^{\alpha}$ so letting $x\to 0$, by continuity of $f$ we get that $f(0)=0$. Thus, the term for $k=0$ is zero. This means that $f_n$ doesn't have any constant terms, since from $k>0$ onwards there are no constant terms, everything is multiplied by at least $x$.

The term for $k=1$ is $nx(1-x)^{n-1} f(\frac 1n)$. The only degree $1$ term in here, from the binomial expansion, is $nxf(\frac 1n)$. But this doesn't contribute very much. What is the reason? Well, $\left|f(\frac 1n)\right| \leq n^{-\alpha}$ so $|nxf(\frac 1n)| \leq n^{1 - \alpha}$ for all $x$.

Finishing the proof

So the final proof is as follows : Let $f$ be a continuous function with $|f(x)| \leq |x|^{\alpha}$ for all $x \in (0,1]$. Let $f_n$ be the $n$th Bernstein polynomial of $f$ and let for $n>1$, $g_n(x) = f_n(x) - nxf(\frac 1n)$. Then $g_n$ has degree at least $2$, and $g_n \to f$ uniformly as $f_n \to f$ uniformly and $nxf(\frac 1n) \to 0$ uniformly in $x$.

Expected extensions

One can probably extend this to the following :

Let $m$ be a natural number and $\alpha>m$ be a real number. Then any $f$ continuous on $[0,1]$ which satisfies $|f(x)| \leq |x|^\alpha$ for all $x \in (0,1]$ can be uniformly approximated by polynomials of the form $a_{m+1}x^{m+1}+...+a_nx^n$ with $n \geq m+1$.

You can try to prove this using the Bernstein polynomial.

Answer 2

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.