Assume $V$ is simply a vector space consisting of real functions $f$ that has $n$th derivative and is 0 outside some bounded real interval. We can make differentiation into $D: V \to V, g \mapsto g'$ and $D^*:V^* \to V^*$ such that $D^*(f) = f \circ D$.
Now, $\phi_{f}(g) = \int_{-\infty}^{\infty} f(x)g(x)dx$. I need to assume that $f =1$ if $x$ is nonnegative and $0$ otherwise. I want to show that $-D^*(\phi_{f}) = \sigma$ where $\sigma(g) = g(0)$
I have tried to solve this by noting that left side is simply just $\int_{-\infty}^{\infty} f(x)g'(x)dx$ and that this can be separated into $\int_{-\infty}^0 f(x)g'(x)dx + \int_0^{\infty} f(x)g'(x)dx$. Then, by $f$'s definition, I simplified this into $C + \int_0^{\infty} g'(x)dx$ and applying boundary condition would generate this to be $C + g(0)$ but I am not sure how to get rid of that constant.
There is also a second part that asks that if we choose $\phi \in V^*$ such that $D^*(\phi) = 0$, then $\phi = \phi_{f}$ where $f(x) = c$ for a real constant $c$. I am simply noting that this would make $\phi_{f} = c\int_{-\infty}^{\infty} g(x)dx$ but I am not sure how to go on from here. Any help would be appreciated. Thanks.