Let $f$ be a continuous function on $[0,1]$. Show that there exists a polynomial $p$ such that $\sup_{x \in[0,1]} |p(x)-f(x)|<\varepsilon$, and $p'(0) = p'(1)= 0$.
The uniform convergence part comes from Weierstrass Approximation Theorem, but how to construct such $p$ to satisfy the condition for derivative?
I think we can start by any polynomial $p_n$ which converges uniformly to $f$, and define a new polynomial $q_n(x) = p_n(x) $ "minus some terms and plus other terms which can be approximated by those terms we subtract" to remain $q_n$ uniformly converging to $f$. But what terms should we subtract? Or are there better ideas? Thanks!
Let $q$ be a polynomial so that $\|f-q\|_{\infty}<\Large\frac{\varepsilon}{3}$. Then consider $$p(x) = q(x) - \frac{q'(1)}{n}x^n + \frac{q'(0)}{n} (1-x)^n,$$
where $n \ge 2$ is large enough so that
$$\bigg| \frac{q'(1)}{n}\bigg| , \bigg| \frac{q'(0)}{n}\bigg| < \frac\varepsilon 3 . $$
Then
$$p'(x) = g'(x) - q'(1) x^{n-1} - q'(0) (1-x)^{n-1}\Rightarrow p'(0) = p'(1) = 0$$
and $\|f - p\|_{\infty} < \varepsilon$.