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.
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.