Is the function an elementary solution?

Question

Is the function : $u : C^{\infty}_0(\mathbb R^2) \rightarrow \mathbb C : f \rightarrow \int_\mathbb R f(t,t) dt$

an elementary ( a fundamental) solution of $\frac{d}{dx_1} + \frac{d}{dx_2}$ ?

I just began the course on distribution theory and I do not know how to start, any comment is welcome.

My attempt so far :

if $u$ is a fundamental solution of $\frac{d}{dx_1} + \frac{d}{dx_2}=:L$ , then $Lu = \delta $ , which means that for any $f \in C^{\infty}_0(\mathbb R^2)$, we have that $<Lu, f>=< \delta, f>= f (0,0)$

But $<Lu, f> = <u, -(\frac{d}{dx_1} + \frac{d}{dx_2}) f> = -(\int_\mathbb R \frac{d}{dx_1}f(t,t) dt + \int_\mathbb R \frac{d}{dx_2}f(t,t) dt)$

Now I have to show that this last term is equal to $f(0,0)$, but I cant find how...

Is my reasoning correct ?

Answer 1

Let's first denote $L=\partial_1+\partial_2$. We have \begin{equation*} \begin{split} -\langle Lu, f\rangle & = \langle u, Lf\rangle = \int_\mathbb{R} (Lf)(t,t) {\,\rm d}t \,. \end{split} \end{equation*} Besides, if $\phi(t)=(t,t)$, we have $\phi'(t)=(1,1)$ and \begin{equation*} \begin{split} (f(t,t))' & = (f\circ\phi(t))' \\ & = (\nabla f)(\phi(t))\cdot\phi'(t) \\ & =(\partial_1 f)(t,t)+(\partial_2 f)(t,t) \\ & = ((\partial_1 f)+(\partial_2 f))(t,t) \\ & = (Lf)(t,t)\,. \end{split} \end{equation*} Hence \begin{equation*} \begin{split} -\langle Lu, f\rangle & = \int_\mathbb{R} (Lf)(t,t) {\,\rm d}t \\ & = \int_\mathbb{R} (f(t,t))' {\,\rm d}t \\ & = f(t,t)\big|_{-\infty}^{\infty} \\ & = 0\,, \end{split} \end{equation*} since ${\rm supp}\,f$ is compact. The conclusion is that $Lu=0$.

Are you sure such $u$ is a fundamental solution for $L$?