Uniform convergence with derivative matters

Question

Suppose I have $C^1$ function $f$ such that both $f$ and $f'$ are bounded, and they can both be approximated uniformly by polynomials. My question is, how can one prove that $\exists\,P_n(x)$ which is a sequence of polynomials such that $P_n(x)\to f$ and $P'_n(x)\to f'$ both uniformly? In other words, I know that $f$ can be approximated uniformly by $P_n(x)$, $f'$ by $p_n(x)$, how is it possible that $P'_n(x)=p_n(x)$? Thanks in advance!

Answer 1

This can be done on a bounded interval $[a,b].$ Let $f\in C^1([a,b]).$ By Weierstrass there is a sequence $p_n$ of polynomials converging uniformly to $f'$ on $[a,b].$ Set $P_n(x) = f(a)+\int_a^xp_n(t)dt.$ Then each $P_n$ is a polynomial, and $P_n'=p_n.$ For any $x\in [a,b]$ we have

$$f(x)-P_n(x) = (f(a)+\int_a^xf') -(f(a)+\int_a^xp_n)= \int_a^x(f'-p_n).$$ Slap absolute values on and we see $$|f(x)-P_n(x)|\le \int_a^b|f'-p_n| \le (b-a)\sup_{[a,b]}|f'-p_n|.$$ The last expression $\to 0$ and is independent of $x.$ Thus $P_n\to f$ uniformly, and of course we had $p_n\to f'$ uniformly all along.