I want to show that if $f\in C([0,1])$ and the distributional derivative $f'$ on $(0,1)$ is in $L^1((0,1))$, then $$f(1) - f(0) = \int_0^1 f'(x)\,dx$$
I am having a lot of trouble getting started. Seems like a combination of using smooth functions and convolutions is probably how I need to go about this, but I'm having trouble coming up with a proof.
On
Let's look at $(f',\varphi)$, where the test function $\varphi$ equals $1$ on $[h,1-h]$, $0\le\varphi\le 1$, and $\varphi$ is supported by $[0,1]$. Since $f'\in L^1$, dominated convergence shows that $(f',\varphi)\to \int_0^1 f'$ as $h\to 0+$.
On the other hand, $$ (f',\varphi)=-(f,\varphi') = -\int_0^h f(x)\varphi'(x)\, dx - \int_{1-h}^h f(x)\varphi'(x)\, dx . $$ We can more specifically fix a smooth $\psi$ that increases from $0$ to $1$ on $[0,1]$ and then take $\varphi(x)=\psi(x/h)$ on $0\le x\le h$. Then the first integral becomes $$ -\frac{1}{h}\int_0^h f(x)\psi'(x/h)\, dx = -\int_0^1 f(ht)\psi'(t)\, dt . $$ By dominated convergence again, this converges to $-f(0)\int_0^1\psi' =-f(0)$ as $h\to 0+$. The second integral can be handled in the same way.
Given $f'\in L^1([0,1])$ let us define a function $$g(x) = \int_0^x f'(t) dt.$$ Now for each $\phi\in \mathcal{D}([0,1])$, if we have $$\int_0^1 [f(x)-g(x)] \phi'(x) dx = 0,$$ this implies that $f - g$ is an a.e. constant function (it is a classic result).
Since $f-g$ is continuous, when we plug in the point $0$ into $f-g$, we see that the constant function is $f(0)$ thus $f(x) = g(x) + f(0)$.