$L^p$ domination of mixed partial derivatives of the 3rd order by the unmixed ones?

Question

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.

Answer 1

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.

Answer 2

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.