Let $V = C_{c}^\infty(\mathbb{R})$ a space of smooth functions on $\mathbb{R}$ with compact support. This basically means the space consists of all functions $f: \mathbb{R} \to \mathbb{R}$ such that the $n$th derivative exists for all $n \ge 1$ and that $f$ is $0$ outside some bounded interval in $\mathbb{R}$. We can denote differentiation as $D: V \to V, g \mapsto g'$. Then, we can define its dual to be $D^*: V^* \to V^*$ such that $D^*(f) = f \circ D$
Define another linear map $\phi_{f}$ such that $\phi_{f}(g) = \int_{-\infty}^\infty f(x)g(x)dx$
I want to show that if the derivative of $f$ exists and is continuous, $-D^*(\phi_{f}) = \phi_{f'}$. I just tried plugging in $\int_{-\infty}^\infty f(x)g(x)dx$ into map $D$ which would get $f(x)g(x)$ and then that would give $f(f(x)g(x))$ but I don't think this is a correct way to do it. (Thanks for the answer, that was such a stupid mistake).
The second part of the question makes $f = 1$ if $x \ge 0$ and $f = 0$ if $x \lt 0$. Then, I am supposed to show that $-D^*(\phi_{f}) = \sigma$ where $\sigma(g)=g(0)$. We know that $f(x)$ is not differentiable. We can see the left side can be split into two intervals: $\int_{-\infty}^0 f(x)g'(x)dx + \int_0^{\infty} f(x)g'(x)dx$. Left term would be 0 because of $f$'s definition so integrating 0 would give you a constant. Right term would just be $\int_0^{\infty} g'(x)dx$ but I am not sure how to progress from here. $\sigma \in V^*$ just to make it clear.
The third part is that if you take $\phi \in V^*$ so that $D^*(\phi) = 0$, then $\phi = \phi_{f}$ where $f(x) = c$ for a real constant $c$. I just have no clue how to do this. Can someone help? Thanks.
This question is about how well you understand the notation, and this means that your approximation was right, just plug the correct objects in the correct equations:
$$D^*(\phi_f)(g)=\phi_f\circ D(g)=\phi_f(g')=\int_0^\infty f(x)g'(x)dx$$
Now you can finish using integration by parts and the "boundary" conditions given.