Show that for a finitely differentiable function there exists a polynomial that both their $k$-th derivatives converge We need to show that for all $f: [a,b] \rightarrow \Bbb R $ which is differentiable $n$ times, and for all $\varepsilon>0$ there exists a polynomial $p\colon[a,b] \rightarrow R$ s.t.
$\forall n \geqslant k\geqslant 0, \forall x\in [a,b] : |f^{(k)}(x)-p^{(k)}(x)|<\varepsilon$
I thought I'd take $p$ as the Taylor series of $f$ and show that it's a polynomial and equals $f$ when the remainder of it converges to zero and then because of equality the derivatives are equal? But I'm really not sure about it... Any leads? :]
 A: We may assume $[a,b]=[-1,1]$; furthermore we assume that $f\in C^n\bigl([-1,1]\bigr)$, where $n\geq0$. The following is a proof by induction.
When $n=0$ (i.e., $f$ is continuous on $[-1,1]$) and an $\epsilon>0$ is given then according to the Stone-Weierstrass theorem we can find a polynomial $p$ such that
$$|f(x)-p(x)|\leq\epsilon\qquad(-1\leq x\leq 1)\ .$$
(Such a polynomial has nothing to do with any Taylor expansion the function $f$ might have.)
Assume now that the statement is true for $n-1\geq0$ and that $f\in C^n\bigl([-1,1]\bigr)$. Applying it to $f'$ we obtain a polynomial $p$ whose derivatives up to order $n-1$ approximate the corresponding derivatives of $f'$:
$$|(f')^{(k)}(x)- p^{(k)}(x)|<\epsilon \qquad(0\leq k\leq n-1)\ .\tag{1}$$
We now define the polynomial $P$ by 
$$P(x):=f(0)+\int_0^x p(t)\ dt\qquad(-1\leq x\leq 1)\ .$$
Then 
$$|f(x)-P(x)|\leq\left|\int_0^x \bigl(f'(t)-p(t)\bigr)\ dt\right|\leq |x|\sup_{-1\leq t\leq1}|f'(t)-p(t)|<\epsilon\ ,$$ and using $(1)$ it is easy to see  that $P$ satisfies stated requirements  for $f$.
Without the assumption that $f^{(n)}$ is continuous the stated claim is false. A sequence of polynomials $(q_k)_{k\geq1}$ converging uniformly to some $g$ on the interval $[-1,1]$ necessitates that $g$ is continuous to begin with. As an example consider the function
$$f(x):=\cases{x^2\sin{1\over x}\quad&$(0<|x|\leq 1)$ \cr 0&$(x=0)$ .\cr}$$
This function is differentiable on all of $[-1,1]$, but as
$$f'(x)=2x\sin{1\over x}-\cos{1\over x}\qquad(x\ne0)$$
there is no polynomial that can approximate $f'$ with an error $<1$ in the $\sup$-norm.
A: I will assume that $f\in C^\infty[a,b]$, other case can be fixed in a similar way. 
Using Stone-Weierstrass theorem and the fact that $f^{(d)}$ is continuous for each $d$, can see that for each $k$, there is a polynomial $P_k$ such that 
$$\max_{0\leqslant d\leqslant k}\max_{x\in [a,b]}|f^{(d)}(x)-P^{(d)}_k(x)|\leqslant \frac 1k.$$
The key is that (when $k=1$), we take $\lVert f'-P\rVert_\infty$ small, then we consider $\int_a^xP(t)dt$, a polynomial.
