# Solving $T' = 0$ for distributions in $\mathbb{R}^n$

Denoting $T \in \mathcal{D}'(\mathbb{R}^n)$ as distributions with $T_f(\varphi) = \int_{\mathbb{R}^n} f\varphi\ dx$, I wish to prove the distribution solution of the equation $T' = 0$ (distribution derivative) is $T = T_c$ where $c$ is constant.
I have a proof for $\mathbb{R}$, where for all $\varphi \in C^\infty_c(\mathbb{R})$ we can have $\varphi = \psi + \varphi_0 T_1(\varphi)$ where $$\psi = \varphi - \varphi_0 T_1(\varphi)$$ choosing $\varphi_0 \in C^\infty_c(\mathbb{R})$ such that $T_1(\varphi_0) = \int_\mathbb{R} \varphi_0 dx =1$. Hence we have $T_1(\psi)= \int_\mathbb{R}\psi dx = 0$, so for $\tau(x) = \int_{-\infty}^x \psi(t)dt$ we have $\psi = \tau'$. Hence if $T' = 0$ then $$T(\varphi ) = T(\tau') + T(\varphi_0)T_1(\varphi) = T(\varphi_0)T_1(\varphi) = T_{T(\varphi_0)}(\varphi)$$ Thus $T = T_{T(\varphi_0)}$. But I can't replicate this on $\mathbb{R}^n$. Any proof for $\mathbb{R}^n$ would be extremely helpful.

• What do you mean by $T'$ in dimension $\geqslant 2$? – Davide Giraudo Mar 11 '13 at 10:42
• I think that $T'$ is some kind of distributional gradient. – Tomás Mar 11 '13 at 13:10
• Everything should be almost the same, take $x=(x_1,\cdots,x_n)$ the integral will be over $\mathbb{R}^n$ and $\tau(x) = \int_{-\infty}^{x_1}\cdots\int_{-\infty}^{x_n} \varphi(t)\mathrm{d}t_1\cdots\mathrm{d}t_n$, $\varphi = \nabla \tau$. – Yimin Mar 11 '13 at 15:26
• @Davide Giraudo In multi dimensions we see the relation of weak derivatives as $D^\alpha T(\varphi) = (-1)^{|\alpha|}T(D^\alpha \varphi)$. So just like gradient $T' = (D^{e_1}T,...,D^{e_n}T)$. In other words I mean that $T'= 0$ implies equivalently $T(\frac{\partial\varphi}{\partial x_i}) = 0$ for all $\varphi \in C^\infty_c(\mathbb{R}^n)$. – smiley06 Mar 12 '13 at 10:59
• @Yimin It doesn't work that way. You are taking $\varphi$ (in the place of $\psi$) as a scalar function. But $\nabla \tau$ is a vector. This is exactly the problem in multi dimensions where to use the decomposition of each $\varphi$ shown above along with the fact that $T(\frac{\partial\tau}{\partial x_i})= 0$ for all $i$ we need to have a $\tau \in C^\infty_c(\mathbb{R}^n)$ such that $\frac{\partial\tau}{\partial x_i} = \psi$ for all $i$. – smiley06 Mar 12 '13 at 11:13