A constant distribution Let $f\in L^{1}_{loc}(\mathbb R^n)$ such that $\nabla T_f = 0$ (i.e., $\frac{\partial T_f}{\partial x_{j}}=0 \quad \forall j=1,2, \dots, n$).
An idea to show that $T_f = \lambda 1$ for some scalar $\lambda$, where $1$ is the indicatrice function.
Thank you in advance
 A: Let $\eta \colon \mathbb{R} \to \mathbb{R}$ be a test function with $\int_{-\infty}^{+\infty} \eta(t)\,dt = 1$. For a test function $\varphi \colon \mathbb{R}^n \to \mathbb{R}$, let $\varphi_0 = \varphi$ and for $k = 1,\dotsc, n$ define $\varphi_k \colon \mathbb{R}^{n-k} \to \mathbb{R}$ by
$$\varphi_k(x_{k+1},\dotsc, x_n) = \int_{-\infty}^{+\infty} \varphi_{k-1}(x_k, x_{k+1}, \dotsc, x_n)\,dx_k,$$
and $\psi_k \colon \mathbb{R}^n \to \mathbb{R}$ by
$$\psi_k(x_1,\dotsc,x_n) = \Biggl(\prod_{j = 1}^k \eta(x_j)\Biggr)\varphi_k(x_{k+1}, \dotsc, x_n).$$
Note that all the $\psi_k$ are test functions, and since
$$\int_{-\infty}^{+\infty} \psi_{k-1}(x) - \psi_k(x)\,dx_k = 0,$$
there is a test function $\chi_k$ with $\psi_{k-1} - \psi_k = \frac{\partial}{\partial x_k}\chi_k$, so
$$T_f[\psi_{k-1} - \psi_k] = T_f[(\partial/\partial x_k) \chi_k] = - \frac{\partial T_f}{\partial x_k}[\chi_k] = 0.$$
Since $\psi_0 = \varphi$ and
$$\psi_n(x) = \prod_{j = 1}^n \eta(x_j)\cdot \int_{\mathbb{R}^n} \varphi(x)\,dx,$$
we thus have
$$T_f[\varphi] = T_f[\psi_n] = T_f\biggl[\prod_{j = 1}^n \eta(x_j)\biggr]\cdot \int_{\mathbb{R}^n} \varphi(x)\,dx.$$
