Show $\{f \in L^2[0,1] : f \text{ is absolutely continuous},\ f\ '\in L^2[0,1],\ f(0)=f(1)=0\}$ is dense in $L^2[0,1]$. Define $D(T)=\{f \in L^2[0,1] : f \text{ is absolutely continuous},\ f\ '\in L^2[0,1],\  f(0)=f(1)=0\}$.
I want to show that $D(T)$ is dense in $L^2[0,1]$.
My argument is as follows:

$\{0,1\} $ is a measure zero set so it suffices to show that absolutely continuous function with derivative in $L^2[0,1]$ is dense in $L^2$. But polynomials on $[0,1]$ are clearly dense (in uniform convergence hence also in $L^2$) by Stone-Weierstrass theorem and are abs. conti. with $L^2$ derivative. Thus, $D(T)$ is dense in $L^2[0,1]$.

Is this a sound argument? Any help is appreciated.
 A: Not all polynomials do belong to $D(T)$ as they do not satisfy the condition $f(1)=0$: Indeed the mapping $f \mapsto f(1)$ is well defined for continuous functions. The measure zero argument does not apply: If $f$ is continuous $f(1)=0$, then $f$ is small in a neighborhoof of $1$, and this neighborhood has no measure zero.
Still $D(T)$ is dense in $L^2$, as it contains $C_c^\infty([0,1])$, the set of
infinitely offen continuously differentiable functions with compact support.
A: Suppose that there exists $f\in L^2$ such that
$$
              \int f(t)g(t)dt = 0
$$
for all $g$ in your set of functions. By a limiting argument, you can easily show that
$$
             \int f(t)\chi_{[a,b]}(t)dt = 0, \;\;\; 0 < a < b \\
             \int_a^b f(t)dt = 0,\;\;\; 0 < a < b
$$
By the Lebesgue differentiation theorem, it follows that $f=0$ a.e.. Hence, the orthogonal complement of your set is $\{0\}$, which means that your set is dense in $L^2[0,1]$.
A: Let $g_{\epsilon} (x)=e^{-\frac {\epsilon} x}e^{-\frac {\epsilon} {1-x}}$ for $0<x<1$ and $0$ for $x \in \{0,1\}$. This function is continuously differentiable (in fact infinitely differentiable). Given $f$ in $L^{2}[0,1]$ first approximate $f$ by a polynomial $p$. Now consider $g_{\epsilon} p$. This function is in $D(T)$ for every $\epsilon$. Also $\|p-pg_{\epsilon}\|_2 \to 0$ as $\epsilon \to 0$ by DCT. [ Note that $0\leq  g_{\epsilon} (x) \leq 1$ for all $x$]. This proves that $D(T)$ is dense. 
