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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.

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    $\begingroup$ What do you mean by $T'$ in dimension $\geqslant 2$? $\endgroup$ – Davide Giraudo Mar 11 '13 at 10:42
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    $\begingroup$ I think that $T'$ is some kind of distributional gradient. $\endgroup$ – Tomás Mar 11 '13 at 13:10
  • $\begingroup$ Everything should be almost the same, take $x=(x_1,\cdots,x_n)$ the integral will be over $\mathbb{R}^n$ and $\tau(x) = \int_{-\infty}^{x_1}\cdots\int_{-\infty}^{x_n} \varphi(t)\mathrm{d}t_1\cdots\mathrm{d}t_n$, $\varphi = \nabla \tau$. $\endgroup$ – Yimin Mar 11 '13 at 15:26
  • $\begingroup$ @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)$. $\endgroup$ – smiley06 Mar 12 '13 at 10:59
  • $\begingroup$ @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 $i$. $\endgroup$ – smiley06 Mar 12 '13 at 11:13

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