I've come upon the result of convolution as
$\langle T_H \star T_H, \psi(x,y) \rangle = \int_0^\infty \int_0^\infty\! \psi(x+y) \, \mathrm{d}x \mathrm{d}y $
and this should be a straighforward application of Fubini's and substitution theorem. The problem is I should be able to express the result as a single distribution (probably regular therefore single integral) and I'm not sure if I'm getting the right boundaries. Consider
$t = x + y , v = x$
Then the Jacobian is equal to $1$ and I think the result should look like
$ \int_0^\infty \int_0^\infty\! \psi(x+y) \, \mathrm{d}x \mathrm{d}y = \int_0^\infty \int_t^\infty\! \psi(t) \, \mathrm{d}v \mathrm{d}t $
Is this wrong? If not, how to proceed further? Thank you
(If there are some confusions about the convolution I'd suggest just focusing on the fubini's theorem. We're using a slightly different definition than what can be seen around the internet)
Well, I figured it out. The last integral is wrong and I should have done this instead
$ \langle T_H \star T_H, \psi(x,y) \rangle =\int_0^\infty \int_0^\infty\! \psi(x+y) \, \mathrm{d}x \mathrm{d}y = \int_0^\infty \int_v^\infty\! \psi(t) \, \mathrm{d}t \mathrm{d}v = \int_0^\infty \int_0^t\! \psi(t) \, \mathrm{d}v \mathrm{d}t = \int_0^\infty t\psi(t) \, \mathrm{d}t = \langle T_{xH},\psi\rangle$