I'm having some trouble with a seemingly simple exercise. I usually know how to verify convergence in the sense of distributions, but this time I'm looking at something a little different. I need to find a distribution $T$ so that $T'\rightarrow 2\delta_1$ in D'(R).
My attempt is the following: using the property of test functions I consider: $$<T',\phi>=-<T,\phi '>=-\int_a^bT\phi 'dx=-T\phi|_a^b+\int_a^bT'\phi dx$$
but the last integral is $<T',\phi>$, so I did this (which I'm not sure is allowed...) $$<T',\phi>=-T\phi|_a^b+<T',\phi> \rightarrow -T\phi|_a^b=0=2\phi(1) \rightarrow T(a)\phi(a)-T(b)\phi(b)=2\phi(1)$$
And now I just arbitrarily decided to consider $a=1 , b=\infty , T(1)=2$ so that I get $$T=2I_{[1,\infty)}$$
where $I_{[1,\infty)}$ is the indicator function of the subset $[1,\infty)$
Proof:
$$<T',\phi>=-<T,\phi '>=-2\int_1^\infty\phi 'dx=-2\[\phi(x)]_1^\infty=-2[-\phi(1)]=2\phi(1)=2\delta_1$$ where $\phi(\infty)=0$ due to test functions being compactly supported.
I'm pretty sure the answer is correct, but the method isn't. I'm not sure how else to tackle this. Thank you.