proving "$C^1([−1,1])$ is dense in the given space with given norm" Define $$E = \left \{ f \in W^{1,2} (-1,1) \; | \; \| f \|_E := \left( \int_{-1}^1 (1-x^2 ) | f' (x) |^2 dx + \int_{-1}^1 | f(x) |^2 dx \right)^{\frac{1}{2}} < \infty   \right \}.$$ Then how can I prove that $ C^1 ([-1,1]) $ is dense in $E$ ? Is this concerned with Sobolev inequality?
 A: I assume that Ann wants the space 
$$E = \left \{ f \in L^2(-1,1) \; \bigg| \; \| f \|_E := \left( \int_{-1}^1 (1-x^2 )
 | f' (x) |^2 dx + \int_{-1}^1 | f(x) |^2 dx \right)^{\frac{1}{2}} < \infty   \right \}.$$
That is, $f\in E$ if it belongs to $L^2(-1,1)$ and  has a weak derivative $f^\prime$ on $(-1,1)$ satisfying
$$ \int_{-1}^1 (1-x^2 ) | f' (x) |^2 dx <\infty.$$ 
Now for $f\in E$, let $p$ be a polynomial with small 
$ \int_{-1}^1 (1-x^2 )\, |p(x)-f'(x)|^2 dx$,
 and then let $q$ be the polynomial with $q(0)=f(0)$ and $q^\prime=p$. 
From the aside below, we see
that $q$ is close to $f$ in $E$-norm, which  shows that polynomials are dense in $E$.

For a more general, multivariable version of this result, see Are polynomials dense in Gaussian Sobolev space?

Aside: Suppose that $f\in E$ and $f(0)=0$.
Then for $0\leq x\leq 1$ we have $f(x)=\int_0^x f^\prime(y)\,dy$. 
Cauchy-Schwarz gives $f(x)^2\leq x\int_0^x (f^\prime(y))^2\,dy,$  and integrating 
in $x$ we find 
$$\int^1_0 f(x)^2\,dx\leq \int_0^1 x \int_0^x (f^\prime(y))^2\,dy\,dx={1\over 2}\int_0^1 (1-y^2) (f^\prime(y))^2\,dy.$$
Adding a similar contribution from negative $x$-values, we conclude that 
$$\int^1_{-1} f(x)^2\,dx\leq {1\over 2}\int_{-1}^1 (1-y^2) (f^\prime(y))^2\,dy.$$
A: $C[-1,1]$ is dense in $L^2[-1,1]$. Any $f \in C[-1,1]$ can be uniformly approximated by a polynomial, hence the polynomials are dense in $L^2[-1,1]$. The polynomials are smooth, hence in $C^1[-1,1]$. It follows that $C^1[-1,1]$ is dense in $L^2[-1,1]$, hence automatically in $E$.
It is not clear to me how this is related to a Sobolev inequality.
