# Writing a distribution as a tensor product

Definition about tensor product of distributions:

Let $$u_j \in \mathcal{D}'(\Omega_j)$$, $$j=1,2$$. There is only $$u \in \mathcal{D}'(\Omega_1 \times \Omega_2)$$ such that $$\langle u, \varphi_1 \otimes \varphi_2 \rangle=\langle u_1, \varphi_1 \rangle \langle u_2, \varphi_2 \rangle$$ $$\forall \varphi_j \in C^{\infty}_{c}(\Omega_j).$$ Furthermore, if $$\varphi \in C^{\infty}_{c}(\Omega)$$ then \begin{align} \langle u, \varphi \rangle&=\langle u_1, \langle u_2, \varphi(x_1,x_2)\rangle \rangle \\ &=\langle u_2, \langle u_1, \varphi(x_1,x_2) \rangle \rangle. \end{align}

My problem is

If $$u \in \mathcal{D}'(\mathbb{R}^{n})$$ and $$x_nu=0$$ then there is $$v\in \mathcal{D}'(\mathbb{R}^{n-1})$$ such that $$u(x)=v(x') \otimes \delta(x_n),$$ where $$x=(x',x_n)$$.

I think I should use the uniqueness of the definition about tensor product.

Take $$\varphi_1 \in \mathcal{D}'(\mathbb{R}^{n-1})$$ and $$\varphi_2 \in \mathcal{D}'(\mathbb{R}).$$
If $$\varphi_2(0)=0$$ then there exists $$\psi \in \mathcal{D}'(\mathbb{R})$$ such that $$\varphi_2(x) = x \, \psi(x)$$ giving $$\langle u, \varphi_1 \otimes \varphi_2 \rangle = \langle u, \varphi_1 \otimes x \psi \rangle = \langle x_n u, \varphi_1 \otimes \psi \rangle = \langle 0, \varphi_1 \otimes \psi \rangle = 0.$$
Otherwise, take $$\rho \in \mathcal{D}'(\mathbb{R})$$ such that $$\rho(0)=1,$$ and set $$\hat{\varphi}_2 = \varphi_2 - \varphi_2(0) \rho.$$ Then $$\hat{\varphi}_2(0) = 0$$ so $$\langle u, \varphi_1 \otimes \hat{\varphi}_2 \rangle = 0.$$ Thus $$\langle u, \varphi_1 \otimes \varphi_2 \rangle = \langle u, \varphi_1 \otimes (\hat{\varphi}_2 + \varphi_2(0) \rho) \rangle = \langle u, \varphi_1 \otimes \hat{\varphi}_2 \rangle + \langle u, \varphi_1 \otimes \varphi_2(0) \, \rho \rangle \\ = \langle u, \varphi_1 \otimes \rho \rangle \varphi_2(0) .$$
Now, define $$v_\rho \in \mathcal{D}'(\mathbb{R}^{n-1})$$ by $$\langle v_\rho, \varphi_1 \rangle = \langle\langle u(x',x_n), \rho(x_n)\rangle, \varphi_1(x')\rangle .$$ Then $$\langle u, \varphi_1 \otimes \rho \rangle \varphi_2(0) = \langle\langle u(x',x_n), \rho(x_n)\rangle, \varphi_1(x')\rangle \, \langle \delta, \varphi_2 \rangle = \langle v_\rho, \varphi_1\rangle \langle \delta, \varphi_2 \rangle = \langle v_\rho \otimes \delta, \varphi_1 \otimes \varphi_2 \rangle.$$
Thus, $$\langle u, \varphi_1 \otimes \varphi_2 \rangle = \langle v_\rho \otimes \delta, \varphi_1 \otimes \varphi_2 \rangle$$ and by uniqueness, we must have $$u = v_\rho \otimes \delta.$$