By the use of the test functions prove that:
$$f(t) \delta '(t) = f(0) \delta '(t) - f'(0) \delta(t)$$
Attempt:
I want to use some test function $\phi (t)$. So starting from
$$\langle f \delta' , \ \phi \rangle = \int^\infty_{-\infty} f(t)\delta'(t) \phi(t) \ dt$$
Using integration by parts:
$$= \Big[ f'(t) \delta (t) \phi (t) \Big]^\infty_{-\infty} - \int^\infty_{-\infty} f'(t) \delta (t) \phi'(t) \ dt$$
I'm stuck here. The aim is to arrive at $\langle f(0) \delta' - f'(0) \delta, \ \phi \rangle$.
I am not sure how to proceed. I think I will need to use the $\langle \delta, \phi \rangle = \phi(0)$ property at some point.
What do I need to do?
Any help is greatly appreciated.
I guess you are using physicist's notations, however the dirac distribution is not associated to a locally-integrable function.
Actually using the definitions, we get quickly the answer. Let $\varphi$ be a test function, one has : $$\langle f\delta', \varphi \rangle=\langle \delta', f\varphi \rangle = -\langle \delta, (f\varphi)' \rangle = -\langle \delta, f'\varphi \rangle-\langle \delta, f\varphi' \rangle= -f'(0)\varphi(0)-f(0)\varphi'(0).$$ But $$f'(0)\varphi(0)=f'(0)\langle\delta,\varphi \rangle$$ and $$f(0)\varphi'(0)=-f(0)\langle \delta',\varphi \rangle,$$ hence $$f\delta' =f(0)\delta'-f'(0)\delta.$$