Polynomials Dense in $C^k([a,b])$ I'm asked to show that  $\mathcal{P} \subset C^{k}([a,b])$ is a dense subset when $C^k([a,b])$ is equipped with the norm
$$||f||_{C^k([a,b])} = \sum_{n=1}^{k} ||f^{(n)}||_{C([a,b]} = \sum_{i=1}^{n} \sup_{x \in [a,b]} |f^{(n)} (x)|.$$ Is the following enough to show this:  
Take $f \in C^k$ and consider $f^{(k)} \in C$.By Weierstrass, there exists a sequence $\{p_{n_1}\}_{n \in \mathbb{N}} \xrightarrow{u} f^{(k)}.$ Now fix $x_0 \in [a,b]$, we may write 
$$f^{(k-1)}(x) = f^{(k-1)} (x_0) + \int_{x_0}^{x} f^{(k)} (t) \, dt.$$
Take a sequence of polynomials $\{p_{k_1}(x_0)\}_{k \in \mathbb{N}} \to f^{(k-1)} (x_0)$ pointwise, which may again be done by Weierstrass. Then, define a sequence $\{p_{n_2}\}_{n \in \mathbb{N}}$ by
 $$p_{n_2} = p_{k_1}(x_0) + \int_{x_0}^{x} p_{n_1}(t) \, dt$$ where $k_1,n_1 \geq n_2$. Then, $p_{n_2}$ is a polynomial with $p^{'}_{n_2} = p_{n_1}$, and $$\{p_{n_2}\}_{n \in \mathbb{N}} \xrightarrow{u} f^{(k-1)}.$$ Induction thus shows that $\mathcal{P}$, the subspace of all polynomials, is dense in $C^{k}([a,b])$.
 A: In order to avoid induction, we may follow the lines of Jackson's theorem (we need something weaker, indeed):

If $f:[-\pi,\pi]\to \mathbb{C}$ is an $r$ times differentiable periodic function such that
  $$|f^{(r)}(x)| \leq 1, \quad -\pi \leq x \leq \pi,$$
  then, for every positive integer $n$, there exists a trigonometric polynomial $T_{n-1}$ of degree at most $n-1$ such that:
  $$|f(x) - T_{n-1}(x)| \leq \frac{C(r)}{n^r}, \quad -\pi \leq x \leq \pi,$$
  where $C(r)$ depends only on $r$.

Sketch of proof: assume that $f\in C^r([-1,1])$ is an even function and let $g(\theta)=f(\cos\theta)$.
Take $g_N(\theta)$ as the truncated Fourier cosine series of $g(\theta)$:
$$ g_N(\theta) = c_0 + \sum_{k=1}^{N} c_k \cos(k\theta),\qquad c_k = \frac{1}{\pi}\int_{-\pi}^{\pi}g(\theta)\cos(k\theta)\,d\theta. $$
By the Riemann-Lebesgue theorem, $c_k=o\left(\frac{1}{k^r}\right)$. Moreover, $\cos(k\theta)$ is bounded by one in absolute value for any $k$ and any $\theta$, hence for every $\varepsilon>0$ there is some $N_\varepsilon$ such that:
$$ \forall N\geq N_\varepsilon,\quad \left\| f - g_N(\arccos x)\right\|_{C^k([-1,1])}\leq \varepsilon $$
and 
$$ c_0 + \sum_{k=1}^{N} c_k\,T_k(x) $$
with $T_k$ being a Chebyshev polynomial of the first kind, is the wanted polynomial approximation.
A: If you use the Weierstrass polynomial approximation theorem as a black-box, I can't imagine a fundamentally different proof than "approximate the $k$'th derivative very well; then compute the $k$'th iterated integral of the approximation (relative to some point in the interval)".  Which is what you did.
