The density of $C^1[0,2\pi]$

Question

I am not sure if the inclusion $\{f \in AC[0,2\pi]: f(0)=f(2\pi)=0\}\subseteq \overline{\{f \in C^1[0,2\pi]: f(0)=f(2\pi)=0\}}.$ Here $C^1[0,2\pi]$ is the set of continuously differentiable functions and $AC[0,2\pi]$ is the set of absolutely continuous functions.

Answer 1

Assuming you mean the closure in the natural topology on $AC$, that being the one given by the norm $|f(0)|+||f'||_1$, then yes. This is clear because any $L^1$ function (of mean zero) can be approximated in $L^1$ by continuous functions (of mean zero).

In detail: Choose a sequence $g_n$ of continuous functions such that $\int_0^{2\pi}g_n=0$ and $||f'-g_n||_1\to0$. Define $$f_n(x)=\int_0^xg_n(t)\,dt.$$Then $f_n\in C^1$, $f_n(0)=f_n(2\pi)=0$, and $f_n\to f$ in the AC norm.