For every continuous function $f:[0,1]\rightarrow \mathbb{R}$, prove that there exists a sequence of polynomials $p_n$ such that $p_n$ converges to $f$ on $[0,1]$ and for every $x\in [0,1]$, we have $p_1(x)<p_2(x)<\cdots.$

My Trial is to use the stone weierstrass theorem, but then i dont know how to "adjust " the polynomials such that they become strictly increasing, please helps.

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    $\begingroup$ Which function does $p_{n+1}(x) - p_n(x)$ approximate? $\endgroup$ – Daniel Fischer Nov 16 '14 at 16:07
  • $\begingroup$ @DanielFischer The zero function...? $\endgroup$ – Jessie Nov 16 '14 at 16:21
  • $\begingroup$ If you let $n\to \infty$. I thought of an arbitrary but fixed $n$. $\endgroup$ – Daniel Fischer Nov 16 '14 at 16:24
  • $\begingroup$ @DanielFischer: the difference between $p_{n+1}-p_n$, so i should let $q_n= p_n+ |p_{n+1}-p_n|$, something like that...? sorry for the dumb question... $\endgroup$ – Jessie Nov 16 '14 at 16:35
  • $\begingroup$ Somewhat similar. You need to construct a sequence of polynomials with the desired properties. In that construction, you use Stone-Weierstraß. If you already have $p1,\dotsc,p_n$, with the desired properties so far, how can you find $p_{n+1}$ continuing the sequence? Well, assume you had such a $p_{n+1}$, let $q_n = p_{n+1}-p_n$, and look at what properties $q_n$ must have. Use S-W to conclude there is such a $q_n$, set $p_{n+1} := p_n+q_n$ and move on to the next step - well, no, actually, formulate it abstractly. $\endgroup$ – Daniel Fischer Nov 16 '14 at 16:43

We just need to show that for any continuous positive function $f$ on $[0,1]$, there exist a polynomial $q(x)$ such that $0<q(x)<f(x)$ holds for any $x\in[0,1]$ and $\|f-q\|_{\infty}$ is arbitrarily small. In such a way, we can build our approximation sequence by taking $p_1(x)$ as any constant stricly smaller than $\min_{x\in[0,1]} f(x)$, then $p_2$ as $p_1+q_1$, $p_3$ as $p_1+q_1+q_2$ and so on.

Let $m=\min_{x\in[0,1]}f(x)>0$. By the Weierstrass approximation theorem, for any $\varepsilon>0$ there exists a polynomial $r(x)$ such that: $$\| f-r\|_{\infty}\leq\varepsilon.$$ Assuming that $\varepsilon<\frac{m}{3}$, we can just take $q(x)$ as $r(x)-2\varepsilon$. In such a way we have: $$ 0<m-\frac{2m}{3}\leq q(x)\leq f(x)+\varepsilon-2\varepsilon < f(x),$$ and: $$\|f-q\|_{\infty}\leq 3\varepsilon $$ that is arbitrarily small, as wanted.

  • $\begingroup$ sorry, but i dont understand what is $q_1, q_2$... $\endgroup$ – Jessie Nov 16 '14 at 16:59
  • $\begingroup$ @chaoli: $q_1$ is the positive polynomial that approximates $f-p_1$ from below. $q_2$ is the positive polynomial that approximates $f-p_1-q_1$ from below and so on. $\endgroup$ – Jack D'Aurizio Nov 16 '14 at 17:01

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