$ \frac{\partial^2 T}{\partial x\partial y} = 0 $, then $ T = ? $

Question

Can we characterize all distributions $T \in \mathcal{D}'(\mathbb{R}^2) $ with the following property of distribution derivatives ? $$ \frac{\partial^2 T}{\partial x\partial y} = 0 $$ For functions it just looks like $ f(x) +g(y) + c $, but having trouble figuring out the general case. Any hints or suggestions are welcome.

Answer 1

Consider two operators from $\mathcal D(\mathbb R^2)$ to $\mathcal D(\mathbb R)$, defined as follows. For a test function $\phi:\mathbb R^2\to\mathbb R$,

• $(I_x \varphi)(y) = \int_{\mathbb R}\varphi(x,y)\,dx$
• $(I_y \varphi)(x) = \int_{\mathbb R}\varphi(x,y)\,dy$

The adjoint of $I_x$ maps $\mathcal D'(\mathbb R)$ into $\mathcal D'(\mathbb R^2)$, which is a fancy way of saying that some distributions $T\in \mathcal D'(\mathbb R^2)$ can be written in the form $S\circ I_x$ for some $S\in \mathcal D'(\mathbb R)$. These are precisely the distributions that satisfy $\dfrac{\partial T}{\partial x}=0$.

Since $\dfrac{\partial T}{\partial y}$ is of the above type, we have $\dfrac{\partial T}{\partial y} = S\circ I_x$ for some $S\in \mathcal D'(\mathbb R)$. Similarly, $\dfrac{\partial T}{\partial x} = R\circ I_y$ for some $R\in \mathcal D'(\mathbb R)$. The distribution $$T- S\circ I_x-R\circ I_y$$ has both $x$- and $y$- derivatives equal to zero, and therefore is constant. The constant can be incorporated into either $S$ or $R$, leading to the answer $$T = S\circ I_x+R\circ I_y$$ with arbitrary $S,R\in \mathcal D'(\mathbb R)$.