Consider a function $g\in L^1(0,1)$ such that $\int_{\frac 12} ^1f' =-\int_0 ^1gf$ for all $f\in C^1(0,1)$ with compact support.
Is such a function $g$ exists? My guess is no. My attempt is as follows:
From the fundamental theorem and from the compactness of the support of every $f$ we know that $f(\frac 12)=\int_0 ^1gf$. Now by choosing $f$ such that $f(\frac 12)=1$ and $\max f(x)=1$ we can derive from the Cauchy-Schwartz inequality that $1\le\int_0 ^1|g|$. Now I want to show that $\int _0 ^1 |g|<1$ but I could not do that. Any ideas?
I do not expect you can show $\int_0^1 |g| <1$. Indeed, the contradiction comes from the conclusion that $\int_0^1 |g|$ is "concentrated at $1/2$":
Note that
$$\tag{1}\lim_{\epsilon \to 0^+} \int_{1/2-\epsilon}^{1/2+\epsilon} |g| = 0 \ \ \text{(How?)}$$
On the other hand, let $f = f_\epsilon$ be in $C^1_0(0,1)$ so that $|f|\le 1$, $f(1/2) = 1$ and $f(y) = 0$ if $|y-1/2|\ge \epsilon$.
Then
$$1 = f(1/2) = \int_0^1 gf = \int_{1/2-\epsilon}^{1/2+\epsilon} gf \le \int_{1/2-\epsilon}^{1/2+\epsilon} |g|.$$
But that will contradict $(1)$ for $\epsilon$ small enough.