Prove that a distribution $F$ is homogeneous of degree $k$ if and only if $\sum_{j=1}^dx_j \frac{\partial F}{\partial x_j}=kF$

Question

Taking derivatives both sides with respect to $a$ in the definition of homogeneous distribution equation $$a^{-d}\int f(x)\phi(\frac{x}{a})=a^k\int f(x)\phi(x)$$ for all $a>0$ and the test function $\phi$ yeilds $$-da^{-d-1}\int f(x)\phi(\frac{x}{a})-a^{-d-2}\int f(x)x\phi'(\frac{x}{a})=ka^{k-1}\int f(x)\phi(x)$$,then taking $a=1$ I get $\sum_{j=1}^dx_j \frac{\partial F}{\partial x_j}=(k+d)F$ instead of $\sum_{j=1}^dx_j \frac{\partial F}{\partial x_j}=kF$ in the sense of distribution.Does anyone know where I'm wrong

Answer 1

You proved that $$(k+d)\langle f,\varphi\rangle = -\langle f ,x\cdot\nabla\varphi\rangle.$$ Now the derivative on the RHS must be brought on $f$: $$\langle f ,x\cdot\nabla\varphi\rangle = \sum_i \left\langle f ,x_i\frac{\partial\varphi}{\partial x_i}\right\rangle = \sum_i \left\langle x_i f ,\frac{\partial\varphi}{\partial x_i}\right\rangle= - \sum_i \left\langle \frac{\partial}{\partial x_i}(x_i f) ,\varphi\right\rangle=$$ $$= - \sum_i \left\langle f+ x_i\frac{\partial f}{\partial x_i} ,\varphi\right\rangle= -d\langle f,\varphi\rangle -\sum_i \left\langle x_i\frac{\partial f}{\partial x_i} ,\varphi\right\rangle.$$ So $$(k+d)\langle f,\varphi\rangle = d\langle f,\varphi\rangle +\sum_i \left\langle x_i\frac{\partial f}{\partial x_i} ,\varphi\right\rangle$$ which implies $$\sum_i x_i\frac{\partial f}{\partial x_i} = kf$$

Answer 2

Only the "only if" part was treated in the original question and the previous answer. Here is the "if" part.

Let $\Phi(a) = F_a(\varphi) = F(\varphi^a)$ for $a>0$, $\varphi \in \mathcal{D}$. It is easy to show that $\Phi(a) \in C^\infty$ for $a>0$ from the continuity of $F$ and the smoothness of $\varphi$.

We have $$ \frac{d\Phi(a)}{da} = \frac{d}{da}F(a^{-d}\varphi_{a^{-1}}) = -\frac{d}{a} F(\varphi^a) - F(a^{-d-2}x_j \frac{\partial \psi}{\partial x_j}), $$ where $\frac{\partial \psi}{\partial x_j}(x) = \frac{\partial \varphi}{\partial x_j}(\frac{x}{a})=a^{d+1}\frac{\partial(\varphi^a)}{\partial x_j}(x)$. So we have $$\frac{d\Phi(a)}{da} = -\frac{d}{a} \Phi(a) - \frac{1}{a}F\left(x_j \frac{\partial (\varphi^a)}{\partial x_j}\right) = -\frac{d}{a} \Phi(a) + \frac{1}{a} \frac{\partial(x_j F)}{\partial x_j}(\varphi^a) = \frac{1}{a} x_j \frac{\partial F}{\partial x_j}(\varphi^a).$$

Now we use the condition $x_j \frac{\partial F}{\partial x_j} = kF$ and get $$\frac{d\Phi(a)}{da} = \frac{k}{a} \Phi(a).$$

This is an easy differential equation. Its solution is $\Phi(a) = a^k \Phi(1)$ or $F_a(\varphi) = F(\varphi^a) = a^k F(\varphi)$.

This concludes the proof.