Let $f \in \mathcal{D}( ${$(x,y) \in \mathbb{R}^2| x>0,y>0)$} and define
$T(\phi(x)):=\int_0^x f(x,y) \phi(y)dy$ for every $\phi \in \mathcal{D}(\mathbb{R})$
Find the adjoint operator $T^*$.
The adjoint operator is defined as: Let $f,g$ be in $\mathcal{D(\mathbb{R})}$ the adjoint operator in relation to T, if existent satisfies the following equation $\langle T(f),g \rangle =\langle f,T^*(g) \rangle$
Let $f \in \mathcal{D}( ${$(x,y) \in \mathbb{R}^2| x>0,y>0)$} further let $\phi,\psi$ be in $\mathcal{D(\mathbb{R})}$
$\langle T(\phi),\psi \rangle =\int_{\Omega} (\int_0^x f(x,y) \phi(y)dy) \psi(x)dx$ where $\Omega \subset \mathbb{R}$
My calculations:
I would like to use Integration by parts for y: $\int_0^x f(x,y)\phi(y)dy=[F \phi]_0^x-\int_0^x F \phi' dy$
I am a little bit confused how to evaluate $[F \phi]_0^x$ , what I did was: $[F \phi]_0^x=F(x,y)\phi(x)-F(0,y)\phi(0)=F(x,y)\phi(x)$
Hope this was correct.
As a result of the integration by parts I get: $\int_0^x f(x,y)\phi(y)dy=F(x,y)\phi(x)-\int_0^x F(x,y) \phi'(y) dy$
$\int_{\Omega} (\int_0^x f(x,y) \phi(y)dy) \psi(x)dx=\int_{\Omega}(F(x,y)\phi(x)-\int_0^x F(x,y) \phi'(y) dy)\psi(x)dx=\int_{\Omega}(F(x,y)\phi(x)\psi(x)-\psi(x) \int_0^x F(x,y) \phi'(y) dy)dx$
It looks like I would need "to get the $\psi$ in to the integral" so that I can continue my calculation. At this point I am stuck.
I would be thankful for a hint.
Here is my calculation: $$ \langle T\phi, \psi \rangle = \int_{-\infty}^{\infty} T\phi(x) \, \psi(x) \, dx = \int_{-\infty}^{\infty} \left( \int_{0}^{x} f(x,y)\,\phi(y) \, dy \right) \, \psi(x) \, dx \\ = \int_{-\infty}^{\infty} \phi(y) \, \left( \int_{y}^{\infty} f(x,y)\,\psi(x) \, dx \right) \, dy = \int_{-\infty}^{\infty} \phi(y) \, T^*\psi(y) \, dy = \langle \phi, T^*\psi \rangle, $$ where $$ T^*\psi(y) = \int_{y}^{\infty} f(x,y)\,\psi(x) \, dx. $$
Have I missed anything caused by the support of $f$ in the first quadrant? I cannot see how that makes a difference. The outer integrals over all of the reals reduce to being over just the positive reals, but as far as I see they can still be written as integrals over all of the reals.