Integration by parts VS fundamental theorem of calculus

Question

Let $f\in BV[-1,1]$ such that $\forall \phi\in C^\infty_c(-1,1)$ continuous functions with compact support, we have $$\int_{-1}^1 f'\phi \ d\lambda=-\int_{-1}^1 f\phi' \ d\lambda$$ I have to prove that $f\in AC[-1,1]$. Of course, thanks to the fundamental theorem of calculus it would be enough to prove that for almost all $x\in (-1,1)$ $$\int_{-1}^x f(s)ds=f(x)-f(-1)$$ but I cannot achieve this result starting only from $C^\infty_c$ functions

Answer 1

Let $F(x)=\int^x_{-1}f'(t)dt$ and let $\phi\in C^{\infty}_c((-1,1)]).$

We have $F(x)\phi(x)=\int^x_{-1}(f'(t)\phi(t)+F(t)\phi'(t))dt.$

Evaluating at $x=1,$ we get $0=\int^1_{-1}(f'(t)\phi(t)+F(t)\phi'(t))dt\Rightarrow$

$\int^1_{-1}(f(t)\phi'(t)-F(t)\phi'(t))dt=\int^1_{-1}(f(t)-F(t))\phi'(t))dt=0$

Now, suppose $w\in C_c^{\infty}(-1,1)$. Then, if $z(x)=\int^x_{-1}w(t)dt$ satisfies $\int^1_{-1}w(t)dt=0$ then $z\in C_c^{\infty}(-1,1)$ and $z'=w$. So, $\int^1_{-1}(f-F)wdt=0$ for all such functions $w$.

But $\overline{C_c^{\infty}(-1,1)}=L^2([-1,1])$ so we have that if $w\in L^2([-1,1])$ satisfies $\int^1_{-1}w(t)dt=0$, then $\int^1_{-1}(f-F)wdt=0$.

Set $K=\int^1_{-1}(f-F)dt$ and $h(t)=f(t)-F(t)-K$. Then, $\int^1_{-1}h(t)dt=0$ so $\int^1_{-1}h(t)(f(t)-F(t))dt=0$ and unwinding the definitions, this implies that $\int^1_{-1}h^2dt=\int^1_{-1}(f-F)hdt-K\int^1_{-1}hdt=0.$ It follows that $f-F-K=0$ a.e. $\lambda.$