2
$\begingroup$

Is it true that for each real $p>1$ there is some real $C_p$ such that for all smooth real-valued functions $u$ compactly supported on $S:=(0,1)^3$ one has $$\|D_1D_2D_3u\|_p\le C_p(\|D_1^3u\|_p+\|D_2^3u\|_p+\|D_3^3u\|_p),$$ where $D_j$ denotes the operator of the partial differentiation with respect to the $j$th argument ($j=1,2,3$) and $\|\cdot\|_p$ is the $L^p(S)$ norm?

The special case when $p=2$ follows easily from the Plancherel isometry.

The similar question for partial derivatives of the 2nd order was answered, affirmatively, in this answer and this answer.

$\endgroup$
1
  • 1
    $\begingroup$ Yes. By the Mikhlin Multiplier Theorem you get an estimate of the form LHS $\lesssim \||D_{1}|^{3}u\|_{L^{p}}+\||D_{2}|^{3}u\|_{L^{p}}+\||D_{3}|^{3}u\|_{L^{p}}$, where $|D_{j}|$ is the operator with symbol $|\xi_{j}|$. Then again by Mikhlin you can replace $|D_{j}|^{3}$ by $D_{j}^{3}$ $\endgroup$ Feb 27 at 19:22

2 Answers 2

7
$\begingroup$

You can do it the way described in the previous answer, no question, but technically, once you knew the result about $D_1D_2$ and $D_1^2,D_2^2$ and also knew that the support does not matter as long as it is compact, you could just observe that if you change $u(x_1,x_2)$ to $u(ax_1,a^{-1}x_2)$, you'll get $$ \|D_1D_2u\|_p^p\le C(a^{2p}\|D_1^2u\|_p^p+a^{-2p}\|D_2^2u\|_p^p)\,, $$
so we actually have $$ \log \|D_1D_2u\|_p\le C+\frac12[\log\|D_1^2u\|_p+\log\|D_2^2u\|_p]\,. $$ In general, the same trick and the 2D second order inequality above (the first one) applied to $D_1^{\alpha_1}\dots D_{i-1}^{\alpha_{i-1}}D_i^{\alpha_i-1}D_{i+1}^{\alpha_{i+1}}\dots D_j^{\alpha_j-1}\dots D_n^{\alpha_n}u$ in the $i,j$ 2D plane with subsequent integration over the other coordinates show that if $\alpha_i,\alpha_j>0$, then $$ \log\|D^\alpha u\|_p\le C+\frac 12[\log\|D^{\alpha'}u\|_p+\log\|D^{\alpha''}u\|_p] $$ where the multiindex $\alpha'$ is obtained from $\alpha$ by adding $(-1,1)$ to $(\alpha_i,\alpha_j)$ and $\alpha''$ is obtained by adding $(1,-1)$.

But then the maximum over $\alpha$ of any given fixed order of the expression $$ \log\|D^\alpha u\|_p+\frac C2\sum_i\alpha_i^2 $$ can be attained only at a pure derivative (when all $\alpha_j$ except one are $0$). End of story.

Of course, this doesn't give you a very good constant, but it is a cheap trickery that you can use without looking in the Grafakos book and such as soon as you know the 2D second order result.

$\endgroup$
5
  • $\begingroup$ Nice trick! Do you know if it has been used elsewhere? $\endgroup$ Mar 7 at 14:15
  • 1
    $\begingroup$ why do all of your answers have some clever trick? $\endgroup$ Mar 7 at 20:44
  • $\begingroup$ @IosifPinelis I'm absolutely sure it has because I learned it somewhere at some point in my life, but if you ask me where exactly I got it from, my memory is blank. $\endgroup$
    – fedja
    Mar 8 at 0:50
  • $\begingroup$ @mathworker21 Maybe it is just all I'm capable of nowadays :lol: $\endgroup$
    – fedja
    Mar 8 at 0:51
  • 1
    $\begingroup$ Nice! Indeed, this is very similar to the usual way to prove Gagliardo-Nirenberg interpolation inequalities. If you write the first inequality in the form $\|D_1D_2u\|\le C\|D_1^2u\|^{1/2}\|D_2^2u\|^{1/2}$ you can work out a similar convexity argument without using logarithms $\endgroup$ Mar 8 at 13:58
4
$\begingroup$

I think similar questions always translate in the $L^p$ boundedness of a Fourier multiplier. In this case you want a Fourier multiplier which "exchanges the operator $D_1D_2D_3$ with the operator $D_1^3+D_2^3+D_3^3$". Consider the Fourier multiplier by the function \begin{equation} m(\xi_1,\xi_2,\xi_3) = \frac{\xi_1\xi_2\xi_3}{|\xi_1|^3+|\xi_2|^3+|\xi_3|^3}. \end{equation} Then as Grafakos, Modern Fourier Analysis, Thrid Edition, Example 6.2.6 notices, when the multiplier has this typ of homogeneity, i.e. in this case \begin{equation} m(\lambda \xi_1, \lambda\xi_2,\lambda\xi_3)=m(\xi_1,\xi_2,\xi_3), \end{equation}by differentiation we get that for any multi index $\alpha \in \mathbb{Z}_{\geq 0}^3$ it holds that \begin{equation} \lambda^{|\alpha|}\partial^\alpha m(\lambda \xi_1, \lambda\xi_2,\lambda\xi_3)=\partial^\alpha m(\xi_1,\xi_2,\xi_3), \end{equation} Then by picking $\lambda = (|\xi_1|^2+|\xi_2|^2+|\xi_3|^2)^{-1/2}$, it follows that \begin{equation} |\partial^\alpha m( \xi_1, \xi_2,\xi_3)| \leq (|\xi_1|^2+|\xi_2|^2+|\xi_3|^2)^{-|\alpha|/2} (\sup_{|\xi|=1}|\partial^\alpha m (\xi)) | \leq c_\alpha |\xi_1|^{-\alpha_1} |\xi_2|^{-\alpha_2} |\xi_3|^{-\alpha_3}. \end{equation} Therefore $m$ satisfies the hypothesis of Corollary 6.2.5 of the same reference, and hence it is a bounded operator on $L^p(\mathbb{R^3}), p>1$. Then you have that \begin{equation} \Vert{D_1D_2D_3} u\Vert_p \leq C_p \sum_{i=1}^3 \Vert |D_i|^3 u \Vert _p =c_p\sum_{i=1}^3 \Vert H_i D_i^3 u \Vert _p \leq c_p \sum_{i=1}^3 \Vert D_i^3 u \Vert _p, \end{equation} where $|D_i|$ is the Fourier multiplier with symbol $|\xi_i|$ and $H_i$ is the Hilbert transform in the $i$-th coordinate.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer. Just one small point: Do you think there is a typo in that Corollary 6.2.5? I think it should be $r\in$ instead of $r\notin$. $\endgroup$ Feb 27 at 19:31
  • $\begingroup$ It seems so, otherwise it seems incompatible with the condition of Theorem 6.2.4. By the way in my answer you need one last step with the one dimensional Hilbert transform in order to pass from the Fourier multiplier $|\xi_i|^3$ to the Fourier multiplier $\xi_i^3$. Do you need to fill in the details ? $\endgroup$ Feb 27 at 19:41
  • $\begingroup$ About the last step, this follows by the $L^p$ boundedness of the Riesz transform, mentioned in your previous answer, right? Anyhow, I think it would be better if you added this detail. $\endgroup$ Feb 27 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.