An $\ell^qL^p$ inequality: $\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{\ell^2L^4}\leq \|{\varphi}\|_{L^2}$.

Question

An $\ell^qL^p$ inequality: $\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{\ell^2L^4}\leq \|{\varphi}\|_{L^2}$. The dimensions I care about are $d=2$ and $d=3$.

We define $\ell^qL^p$ in the following way. For $\alpha\in \mathbb Z^d$, let $\square_\alpha$ be the unit cube in $\mathbb R^d$ with centre at $\alpha$ and let $\chi_\alpha$ be its characteristic function.

Define $\ell^q(L^p)$ to be the set of functions for which $$\|{f}\|_{\ell^qL^p}:=\left({\sum_\alpha\|{f\chi_\alpha}\|^q_{L^p}}\right)^{\frac{1}{q}}=\left({\sum_\alpha\left({\int_{\square_\alpha}|{f(x)}|^p\mathrm{d}x}\right)^{\frac{p}{q}}}\right)^{\frac{1}{q}}$$is finite.

The question says to use the Hölder inequality and the Sobolev inequality.

Idea of proof:

Using $L^p$ interpolation, $\frac{1}{4}=\frac{\frac{1}{2}}{2}+\frac{\frac{1}{2}}{\infty}$, so \begin{align*}\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{\ell^2L^4}= \left({\sum_\alpha\|{\nabla\langle\nabla\rangle^{-2}\varphi\chi_\alpha}\|^2_{L^4}}\right)^{\frac{1}{2}}&\leq\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{L^\infty}^{\frac{1}{2}}\left({\sum_\alpha\|{\nabla\langle\nabla\rangle^{-2}\varphi\chi_\alpha}\|^2_{L^2}}\right)^{\frac{1}{2}}\\ &= \|\nabla\langle\nabla\rangle^{-2}\varphi\|_{L^\infty}^{\frac{1}{2}}\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{L^2}^{\frac{1}{2}}.\end{align*}

Now $\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{L^2}\leq\|\varphi\|_{L^2}$ which is easily seen on the Fourier side. I'm not sure how to prove $\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{L^\infty}$ is bounded by $\|\varphi\|_{L^2}$ (I'm not even sure if this is true).

Answer 1

By defining $\psi := (1-\Delta)^{-1}\varphi = \langle\nabla\rangle^{-2}\varphi$, then $\varphi = (1-\Delta)\psi$ and your inequality can be written $$ \|\nabla\psi\|_{\ell^2L^4} \leq \|(1-\Delta)\psi\|_{L^2}. $$ Let me first take $d\geq 3$ and $p = \frac{2\,d}{d-2}$. Since $2\leq 4\leq p$, by Hölder's inequality, for any unit cube $\square$ $$ \|\nabla\psi\|_{L^4(\square)} \leq \|\nabla\psi\|_{L^2(\square)}^{\theta} \|\nabla\psi\|_{L^p(\square)}^{1-\theta} $$ where $\theta = \frac{p-4}{2(p-2)}$ (i.e. $\frac{1}{4} = \frac{\theta}{2}+\frac{(1-\theta)}{p}$). Then the Sobolev inequality tells that $$ \|\nabla\psi\|_{L^p(\square)} \leq C_d\,\|\nabla\psi\|_{H^1(\square)}, $$ and so we deduce that $$ \|\nabla\psi\|_{L^4(\square)} \leq C_d\,\|\nabla\psi\|_{L^2(\square)}^{\theta} \|\nabla\psi\|_{H^1(\square)}^{1-\theta} \leq C_d\, \|\nabla\psi\|_{H^1(\square)}. $$ Taking the squares on both sides, then the sum over all the cubes and then the square root yields $$ \|\nabla\psi\|_{\ell^2L^4} \leq C_d\, \|\nabla\psi\|_{H^1}. $$ It remains to notice that $$ \|\nabla\psi\|_{H^1}^2 \leq \|(1-\Delta)\psi\|_{H^1}^2 $$ as can be easily seen from the Fourier definition of $H^1$ for example. In dimension $d\leq 2$, the Sobolev inequality still works from $H^1$ to any $L^p$ with $p\in[2,\infty)$.