If $p,q$ are distributions with $\partial_ip=\partial_iq$, then $p=q$

Question

Let

• $d\in\mathbb N$
• $\Omega\subseteq\mathbb R^d$
• $\mathcal D(\Omega):=C_c^\infty(\Omega)$

Let $$\frac{\partial p}{\partial x_i}(\phi):=-p\left(\frac{\partial \phi}{\partial x_i}\right)\;\;\;\text{for }\phi\in\mathcal D(\Omega)$$ for $p\in\mathcal D'(\Omega)$. Is the mapping $$\frac\partial{\partial x_i}:\mathcal D'(\Omega)\to\mathcal D'(\Omega)\tag 1$$ injective?

At a first glance, I wouldn't think so, cause $\frac\partial{\partial x_i}:C_c^1(\Omega)\to C_c^0(\Omega)$ is not injective, but on the other hand each $\phi\in\mathcal D$ can be written as the $i$-th partial derivative of some $\psi\in\mathcal D(\Omega)$ ...

Answer 1

No, this map is not injective. Indeed, the inclusion $C^\infty(\Omega)\to \mathcal{D}'(\Omega)$ mapping a function to the regular distribution it generates is injective and the $i$-th partial derivative of a regular distribution generated by some $f\in C^\infty(\Omega)$ is the regular distribution generated by $\partial_i f$. So if $f,g\in C^\infty(\Omega)$ have differ only by a constant, the regular distributions they generate have the same partial derivatives.

Your first argument fails for $d=1$: For an interval $I\subset\mathbb{R}$, the derivative is injective on $C_c^1(I)$. But there is no need to restrict to compactly supported functions. However, in principle you had the right idea how to proceed.

Your second assertion is false. Let's restrict to $d=1$ for simplicity: If $\phi\in C_c^\infty((a,b))$, then $\int_a^b\phi'\,dx=\phi(b)-\phi(a)=0$. But it is certainly not true that $\int_a^b \psi\,dx=0$ for every $\psi\in C_c^\infty((a,b))$. In particular, a positive function $\psi\in C_c^\infty((a,b))$ is the derivative of a function $\phi\in C_c^\infty((a,b))$ if and only if it is zero everywhere.

Answer 2

For simplicity, assume that $i=d$ and $d \ge 2$. Let $u \in \mathcal D ' (\Bbb R ^{d-1})$. View it as a distribution on $\Bbb R^d$ by tensorizing it with $1$, i.e. $\langle u \otimes 1, \varphi \rangle = \langle u, \int \varphi (\cdot, x_d) \ \Bbb d x_d \rangle$. Then

$$\langle \partial _{x_d} (u \otimes 1) , \varphi \rangle = - \left< u, \int \partial _{x_d} \varphi (\cdot, x_d) \ \Bbb d x_d \right> = - \langle u, \varphi (\cdot, + \infty) - \varphi (\cdot, -\infty) \rangle = - \langle u, 0, \rangle = 0$$

for all $\varphi \in \mathcal D (\Bbb R^d)$, so $\partial _{x_d} (u \otimes 1) = 0$ even though $u$ may chosen not to be $0$, which shows that the partial derivative operator does not act injectively.

If $d=1$, consider all the constant functions viewed as distributions - they are all mapped to $0$ by an argument as above (which only uses the Leibniz-Newton formula together with the fact that $\varphi$ vas compact support), so again the mapping is not injective.