I am working on the following problem trying to use strategy in this problem. I am trying to simplify the proof by working with $v_{i}=0,i\not=1$ case. But the result looks very different from what I expected, so I want to ask if there is something wrong in my computation. To be more precise, I do not understand why $(1-n)$ term appeared in my computation and my result may diverge.

Let $v_{1},...,v_{n}$ be functions of class $\mathbb{C}^{1}(\mathbb{R}^{n}-\{0\})$ such that $$v_{i}(xt)=t^{1-n}v_{i}(x)$$ for $0\not=x\in \mathbb{R}^{n}$, $t>0,i=1,...n$. Show that, for $\phi\in C^{\infty}_{c}(\mathbb{R}^{n})$ we have $$\sum^{n}_{i=1}\langle \partial_{i}v_i,\phi\rangle=PV \int \phi\sum^{n}_{i=1}\partial_{i}v_{i}dx+\phi(0)\int_{\mathbb{S}^{n-1}}\sum^{n}_{i=1}\theta_{i}v_{i}(\theta)d\omega(\theta)$$

This obviously follows if we can prove the one dimension claim by letting $v_{2}..v_{n}=0$: $$\langle \partial_{1}v_{1},\phi\rangle=PV \int \phi[\partial_{1}v_{1}]dx+\phi(0)\int_{\mathbb{S}^{n-1}}\theta_{1}v_{1}(\theta)d\omega(\theta)$$ And this follows if and only if we have $$\lim_{\epsilon\rightarrow 0}\int^{|x|\le \epsilon}_{|x|\ge 0}[\partial_{1}v_{1}]\phi dx=\phi(0)\int_{\mathbb{S}^{n-1}}\theta_{1}v_{1}(\theta)d\omega(\theta)$$ We know $x=r\theta$, thus we have $$\frac{\partial r}{\partial x_{1}}=\frac{x_{1}}{r}=\theta_{1}$$ Therefore(in the following we no longer distinguish $v_{1}$ and $v$) \begin{align*} \frac{\partial}{\partial x_{1}}v(r\theta) &=\frac{\partial}{\partial x_{1}}[r^{1-n}v(\theta)]\\ &=r^{1-n}\frac{\partial}{\partial x_{1}}v(\theta)+[\frac{\partial}{\partial x_{1}}r^{1-n}]v(\theta)\\ &=r^{1-n}\frac{\partial}{\partial x_{1}}v(\theta)+(1-n)r^{-n}\frac{\partial r}{\partial x_{1}}v(\theta)\\ &=r^{1-n}\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)r^{-n}\frac{x_{1}}{r}v(\theta)\\ &=r^{1-n}(\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)\frac{\theta_{1}}{r} v(\theta)) \end{align*} Substituting $x=r\theta$ we now have: \begin{align*} \lim_{\epsilon\rightarrow 0}\int^{|x|\le \epsilon}_{|x|\ge 0}[\partial_{x_{1}}v]\phi dx &=\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}r^{1-n}(\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)\frac{\theta_{1}}{r} v(\theta))\phi(r\theta)drdsr^{n-1}\\ &=\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}(\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)\frac{\theta_{1}}{r} v(\theta))\phi(r\theta)drds\\ &=\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}dr\int_{\theta\in \mathbb{S}^{n-1}}\frac{\partial v(\theta)}{\partial x_{1}}ds\\ &+(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}\frac{\phi(r,\theta)}{r} v(\theta)\theta_{1}drds\\ &=(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}\frac{\phi(r,\theta)}{r} v(\theta)\theta_{1}drds \end{align*} Let $\phi(r,\theta)$ have the Taylor expansion around 0: $$\phi(r,\theta)=\phi(0,\theta)+r\psi(r,\theta)$$ with $\psi=\frac{\int^{r}_{0} \phi'(t,\theta)dt}{r}$. Then the above integral become \begin{align*} (1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}[\frac{\phi(0)}{r}+\psi(r,\theta)]v(\theta)\theta_{1}drds \end{align*} Note this integral diverges if $\phi(0)\not=0$: $$(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}[\psi(r,\theta)]v(\theta)\theta_{1}drds+(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\frac{\phi(0)}{r}\int_{\theta\in \mathbb{S}^{n-1}}v(\theta)\theta_{1}drds$$

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    $\begingroup$ You have not factored in the singularity at the origin for $\partial_1 v_1$. In other words, the singularity you removed by doing the principle value computation got added back again somewhere in your computation. $\endgroup$ Jun 19, 2012 at 9:17

1 Answer 1


The main problem with your derivation is that your assertion

$$ \lim_{\epsilon \to 0} \int_{0 < |x| < \epsilon} \partial_1 v_1(x) \phi(x) \mathrm{d}x = \langle \partial_1 v_1,\phi\rangle - PV \int \partial_1 v_1 \phi \mathrm{d}x $$

is not true. In other words, while away from the origin the distribution $\partial_1 v_1$ can be identified with the function $\partial_1 v_1$, at the origin (which you omit in your formulation of the expression of the LHS) the two are not equal (mainly because the derivative of $v_1$ is expected to scale like $r^{-n}$ near the origin and is not locally integrable at the origin, so you cannot expect the distributional derivative $\partial_1 v_1$ to be represented by a function), which is why you end up with an extra blow-up.

The correct derivation uses the definition of a derivative of a distribution, that is

$$ -\langle \partial_1 v_1, \phi\rangle = \langle v_1,\partial_1\phi\rangle = \int v_1 \partial_1 \phi \mathrm{d}x = \int_{|x|<\epsilon} v_1\partial_1\phi\mathrm{d}x + \int_{|x| > \epsilon} v_1 \partial_1 \phi \mathrm{d}x \tag{1}$$

Consider first the interior integral, using homogeneity of $v_1$ we have, in polar coordinates (using that $v_1$ is a locally integrable function, so the change of variables is allowed and doesn't do anything weird at the origin)

$$ \left|\int_{|x| < \epsilon} v_1 \partial_1\phi \mathrm{d}x\right| = \left| \int_{\mathbb{S}^{n-1}}\int_0^\epsilon v_1(\theta) r^{1-n} (\partial_1\phi)(r\theta) r^{n-1}\mathrm{d}r\mathrm{d\theta}\right| \leq \left\|\mathbb{S}^{n-1}\right\| \cdot \sup_{\mathbb{S}^{n-1}}\left|v_1(\theta)\right| \cdot \sup_{\mathbb{R}^n} \left|\partial_1\phi(x)\right|\cdot \epsilon $$

which means that the first term is $O(\epsilon)$ and vanishes as $\epsilon \to 0$.

The second term we integrate by parts (or, in other words, reverse Leibniz rule)

$$ \int_{|x|\geq \epsilon} \sum_{i = 1}^nv_i \partial_i\phi\mathrm{d}x = \int_{|x|\geq \epsilon} \sum_{i = 1}^n \partial_i\left( v_i\phi\right)\mathrm{d}x - \int_{|x|\geq \epsilon} \sum_{i = 1}^n (\partial_i v_i ) \phi\mathrm{d}x $$

and treat the first term on the RHS by the divergence theorem

$$ \int_{|x| \geq \epsilon} v_1 \partial_1 \phi\mathrm{d}x = - \int_{|x|\geq \epsilon} \partial_1 v_1 \phi \mathrm{d}x - \int_{\mathbb{S}^{n-1}} v_1(\epsilon\theta) \theta_1 \phi(\epsilon\theta) \epsilon^{n-1}\mathrm{d}\theta \tag{2}$$

the $-\theta_1$ we pick up when we apply divergence theorem which requires a dot product with the "outward pointing normal" against the vector $(v_i)$. Since $v_2 \ldots v_n = 0$ the dot product simplifies to $-\theta_1 v_1$, using that we are outside the disc of radius $\epsilon$ so the "outward pointing normal" actually points toward the origin.

Now the first term in the RHS of (2) is the term used to define the PV integral, so combining (1) and (2) and using that the first term on the RHS of (1) is $O(\epsilon)$ we have that your claim would be proved if we can show that

$$\lim_{\epsilon\to 0} \int_{\mathbb{S}^{n-1}} v_1(\epsilon\theta) \theta_1 \phi(\epsilon\theta) \epsilon^{n-1}\mathrm{d}\theta = \phi(0) \int_{\mathbb{S}^{n-1}} v_1(\theta)\theta_1 \mathrm{d}\theta $$

But this follows immediately by continuity of $\phi$ at the origin and the homogeneity of $v_1$.

  • $\begingroup$ I am confused with the last step. Are you using integration by parts and the divergence theorem in the same time? I feel confused.. $\endgroup$ Jun 19, 2012 at 19:50
  • $\begingroup$ Which step do you mean? The very last step is just a continuity argument: for every $\delta$ there exists an $\epsilon$ such that $|\phi(\epsilon\theta) - \phi(0)| < \delta$ for all $\theta$. Hence by choosing $\epsilon$ sufficiently small the left and right hand sides of the (unnumbered) last equation can be made arbitrarily close to each other. If you mean the step leading up to equation (2), let me make an edit. $\endgroup$ Jun 20, 2012 at 7:46
  • $\begingroup$ Basically, when you integrate by parts in one dimension, you get boundary terms: those come from the use of fundamental theorem of calculus on the total derivative term. When you integrate by parts in higher dimensions, you get boundary terms, these should be interpreted as applying the divergence theorem. Note that the divergence theorem can be derived as a consequence of the fundamental theorem of calculus, and that the fundamental theorem of calculus can be seen as a special case of the divergence theorem. $\endgroup$ Jun 20, 2012 at 7:52

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