Weierstrass approximation theorem application?

Question

I want to prove that if $f,g \in C([0,1]), f(x)<g(x) \forall x \in [0,1]$ then exists a polynomial $P(x)$ such that $$f(x)<P(x)<g(x) \forall x \in [0,1]$$

My work:

Since $f,g \in C([0,1])$ then by Weierstarss approximation theorem exists a sequence of polynomials $P_n$ such that $$\lim_{n \rightarrow \infty} P_n^f(x) = f(x)$$ $$\lim_{n \rightarrow \infty} P_n^g(x) = g(x)$$ uniformly on $[0,1]$

Also, let $\lim_{n \rightarrow \infty} P_n^f(x) = P_n - \varepsilon$ and $\lim_{n \rightarrow \infty} P_n^g(x) = P_n + \varepsilon$, by the same Theorem.

Hence, $f(x) = \lim_{n \rightarrow \infty} P_n^f(x) < P_n < \lim_{n \rightarrow \infty} P_n^g(x) = g(x)$

Q.E.D.?

Answer 1

Hint: You do not need to find a polynomail which converges to $f$ or $g$. Indeed, let $h$ be another continuous function so that

$$f(x) < h(x) < g(x).$$

Then use the theorem to find a polynomial $P$ which is closed to $h$. If $P$ and $h$ are closed enough, you still have

$$f(x) < P(x) < g(x).$$