$L^2$ limit of $C^1$ functions Let $\{g_n\}$ be a sequence of $C^1[0,1]$ functions such that $g_n(0) = g_n(1) =0$. Is it possible to determine any conditions on the $L^2$-limit of $\{g_n'\}$? 
Suppose $g_n \to h$ in $L^2$. Then \begin{eqnarray}
\left( \int_0^1 \left| g_n(x) - h(x) \right|^2 dx\right)^{1/2} \to 0, \ n \to \infty.
\end{eqnarray} But $L^2$ convergence doesn't give us much. We can't assert that $h$ is $C^1$, I'm thinking that $h$ is possibly absolutely continuous, but can't prove it. 
 A: You might consider the Cantor function as your $h$. Then there exists a sequence $\{g_n\}$ of $C^1[0,1]$ functions that converges to $h$ uniformly, thus also in $L^2$ (the $\{g_n\}$ could be a smooth version of the standard continuous piecewise linear approximation to the cantor function, simply by smoothifying at the points where the piecewise linear functions are not differentiable). But the Cantor function $h$ is not absolutely continuous. 
A: The functions $\{ e_n(x)=\sqrt{2}\sin(n\pi x) \}_{n=1}^{\infty}$ form an orthonormal basis of $L^2[0,1]$. So every $f\in L^2$ can be written as the $L^2$ convergent series
$$
           f = \sum_{n=1}^{\infty}\langle f, e_n \rangle e_n.
$$
The sequence $\{ g_n\}_{n=1}^{\infty}$ defined by $g_n = \sum_{k=1}^{n}\langle f,e_n\rangle e_n$ consists of functions in $C^{\infty}[0,1]$ which vanish at $0,1$. So this is a really nice set of functions. These functions are also real analytic. You can't get must nicer than that. And, yet, $\{ g_n \}_{n=1}^{\infty}$ converges in $L^2[0,1]$ to $f$, for an arbitrary $f\in L^2[0,1]$. There's a subsequence that will converge pointwise a.e. to $f$ because $\{ g_n \}$ converges in $L^2$ to $f$.
If $f$ is absolutely continuous, then $f'\in L^1$ and
\begin{align}
         \int_{0}^{1}f(x)e_n(x)dx &= \sqrt{2}\int_{0}^{1}f(x)\sin(n\pi x)dx \\
        & = \left.-\sqrt{2}f(x)\frac{\cos(n\pi x)}{n\pi}\right|_{x=0}^{1}
   +\frac{\sqrt{2}}{n\pi}\int_{0}^{1}f'(x)\cos(n\pi x)dx,
\end{align}
which leads to the bound
$$
        | \langle f,e_n\rangle | \le  \frac{C}{n},\;\; C=\frac{\sqrt{2}}{\pi}\left(|f(0)|+|f(1)|+\int_{0}^{1}|f'(x)|dx\right).
$$
However, $f=\sum_{n=1}^{\infty}\frac{1}{n^{0.51}}e_n \in L^2$, and does not satisfy the above condition, which leads to the sequence $\{ g_n = \sum_{k=1}^{n}n^{-0.51}e_n\}_{n=1}^{\infty}\subset C^{\infty}[0,1]$ satisfying $g_n(0)=g_n(1)=0$ that converges in $L^2$ to a function $f\in L^2$ that is not absolutely continuous. The sequence of derivatives $\{ g_n' \}$ cannot converge in $L^2[0,1]$ to some $h$ because that would force $f$ to be absolutely continuous, as one would have
$$
          \int_{0}^{x}g_n(u)du = \int_{0}^{x}\int_{0}^{u}g_n'(v)dv \\
      \int_{0}^{x}f(u)du = \int_{0}^{x}\int_{0}^{u}h(v)dvdu \\
         f(x) = \int_{0}^{x}h(v)dv.
$$
