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I am trying to understand the following paper. In page 1191, in the beggining of the proof of Theorem 2.9. the authors consider the convolution $$v=\Gamma\star f$$

They claim that $v\in W^{1.p}(\mathbb{R}^n)$. Does anyone knows why this is true with $f$ being only in $W^{-1,p}(\mathbb{R}^n)$?

Update:

Maybe this can thrown some light upon the matter.

First we have that $f\in W^{-1,p}(\Omega)$. This implies the existence of $F\in L^{p'}(\Omega)$ such that $$\langle f,\varphi\rangle=\int_\Omega F\cdot\nabla\varphi,\ \forall\ \varphi\in W_0^{1,p}(\Omega) $$

Extend $F$ by zero outside $\Omega$, so $f\in W^{-1,p}(\mathbb{R}^3)$ and $$\langle f,\varphi\rangle=\int_\Omega F\cdot\nabla\varphi,\ \forall\ \varphi\in W_0^{1,p}(\mathbb{R}^3)$$

By definition $\Gamma\star f$ is a tempered distribution defined by $$\langle\Gamma\star f,\varphi\rangle=\langle \Gamma,\langle f,\varphi(x+\cdot)\rangle\rangle,\ \forall\ \varphi\in \mathcal{S}$$

Note that \begin{eqnarray} \langle\Gamma\star f,\varphi\rangle &=& \int_{\mathbb{R}^3}\Gamma(x)\int_{\mathbb{R}^3}F(y)\cdot\nabla\varphi(x+y)dydx \nonumber \\ &=& \int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\Gamma (z-y)F(y)\cdot\nabla\varphi(z)dzdy \nonumber \\ &=& \int_{\mathbb{R}^3}\left(\Gamma\star F\right)(z)\cdot\nabla\varphi(z)dz \\ &=& -\int \operatorname{div}(\Gamma\star F)(z)\varphi(z)dz \\ &=& \left\langle -\sum_{i=1}^3\frac{\partial\Gamma}{\partial x_i}\star F,\varphi\right\rangle,\ \forall\ \varphi\in\mathcal{S} \end{eqnarray}

If my calculations are right, we can see $\Gamma\star f$ as the functions defined by that $$(\Gamma\star f)(x)=-\sum_{i=1}^3\left(\frac{\partial\Gamma}{\partial x_i}\star F\right)(x)$$

Are my calculations right? If so, can we derive the desired regularity from the above expression?

Thank you

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For $f \in W^{-1,p}$, I take it you mean the dual of $W^{1,q}$, which is to say that it is a bounded linear operator on functions lying in $W^{1,q}$ where $q$ is the conjugate Holder exponent of $p$.

Now let's consider the action of $f$ as a distribution, which is to say, $f$ acts on test functions $\psi$ in a way that is linear and bounded by the $W^{1,q}$ norm of $\psi$. Using the $L^p$ duality to $L^q$, we conclude that there are functions $F_i$ and $F$ such that $$ \langle f, \psi \rangle = \sum_i \int F_i \frac{\partial \psi}{\partial x_i} $$ just as you have already done (since the $W^{1,q}$ norm of $\psi$ and the $L^q$ norm of $\nabla \psi$ are equivalent due to the Sobolev embedding). Furthermore, $$\langle f,\psi \rangle \leq \|f\|_{-1,p} \|\psi\|_{1,q}$$

Using the usual rules, we see that $$ \langle f * \Gamma, \psi \rangle = \sum_i \int F_i(y) \Gamma(x - y) \frac{\partial \psi}{\partial x_i}(x) dx dy = \langle f, \Gamma*\psi\rangle$$ To find the $L^p$ norm of $f*\Gamma$, we take $$ \|f*\Gamma\|_p = \sup_{\|\phi\|_q = 1} \langle f*\Gamma, \phi \rangle \leq \sup_{\|\phi\|_q = 1} \|f\|_{-1,p} \|\Gamma * \phi\|_{1,q} \leq C_{n,q} \|f\|_{-1,p}$$ where at the last step we have used the Hardy-Littlewood-Sobolev inequality, that $\|\psi * \Gamma\|_{2,q} \leq C_{n,q} \|\psi\|_q$.

Similarly, to find the $L^p$ norm of $\frac{\partial \Gamma * f}{\partial x_i}$, we take $$ \|\frac{\partial f*\Gamma}{\partial x_i}\|_p = \sup_{\|\phi\|_q = 1} \langle \frac{\partial f*\Gamma}{\partial x_i}, \phi \rangle = \sup_{\|\phi\|_q = 1} \langle f*\Gamma,\frac{\partial \phi}{\partial x_i} \rangle = \\ \sup_{\|\phi\|_q = 1} \sum_j \int F_j(y) \Gamma(x-y) \frac{\partial^2 \phi}{\partial x_i \partial x_j}(x) dx \leq \\ \|f\|_{-1,p} \sup_{\|\phi\|_q = 1} \|\Gamma * \phi\|_{2,q} \leq C_{n,q} \|f\|_{-1,p}$$

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  • $\begingroup$ Very interesting, thank you. Just one question: Is my identification: $$(\Gamma\star f)(x)=-\sum_{i=1}^3\left(\frac{\partial\Gamma}{\partial x_i}\star F\right)(x)$$ correct? $\endgroup$ – Tomás Jul 1 '13 at 13:04
  • $\begingroup$ Yes, I think it is. However, you will notice the peculiar thing - the precise identification of the distribution with particular integrals is not necessary for the proof - if I had wanted to be more abstract, I could have written it purely in terms of norms and operations. $\endgroup$ – Ray Yang Jul 1 '13 at 13:14
  • $\begingroup$ I agree with you. I just want to make sure if my calculations are right. Thank you again. $\endgroup$ – Tomás Jul 1 '13 at 13:29

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