Let $C_c^{\infty}(\mathbb{R})$ be the space of compactly supported smooth functions on $\mathbb{R}$, we define a functional $\phi_f(g)= \int_{\mathbb{R}} fg dx$ on $C_c^{\infty}(\mathbb{R})$ and let $D$ be the differentiation operation on $C_c^{\infty}(\mathbb{R})$.
How to prove that $D'\phi_f = \phi_{-Df}$?
How to show that if $D'\phi = 0$ then $\phi = \phi_{c}$ for some constant $c$?
Best Answer
$\phi_{-Df}(g)=\int_\mathbb{R}-Dfgdx=\int_\mathbb{R}D(fg)dx-\int_\mathbb{R}Dfgdx=$
$=\int_\mathbb{R}fD(g)dx=\phi_f(Dg)$
While if $D’\phi_f=0$ then
$\phi_{-Df}(g)=-\int_\mathbb{R}D(f)gdx=0$
for each $g$
In general when you have a $C^\infty_c(\mathbb{R})$ function $h$ such that
$\int_\mathbb{R}hgdx=0$ for each $g$ then
$h=0$
In fact you can choose $g=h$ to have
$\int_\mathbb{R}h^2dx=0$ that implies $h=0$
So in your case $D(f)=0$ that means $f=c$
You can observe that $c=0$ because $f$ is continuos with compact support.