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

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).

• What do you mean exactly by $\langle \nabla \rangle^{-2} \varphi$ here?
– cs89
Apr 27 at 20:28
• It is usually a notation for $(1-\Delta)^{-1}\varphi$. I suppose you should use the fact that by Sobolev inequality $\|\nabla\langle\nabla\rangle^{-2}\varphi\|_{L^p(\square)} \leq\|\varphi\|_{L^2}$ where $p = 2^* > 2$, so you should rather interpolate between $L^2$ and $L^p$. Apr 27 at 22:18
• So $\psi = \langle \nabla \rangle^{-2} \varphi$ is the solution to $(1-\Delta)\psi = \varphi$ on $\mathbb{R}^d$. But then there is little hope to write local inequalities. Like, do you intend to use the $L^2(\square)$ norm on the RHS in your estimate?
– cs89
Apr 28 at 5:59
• @cs89 Define $\langle\nabla\rangle^{-2}$ to be an operator on $L^2$ such that $\widehat{\langle\nabla\rangle^{-2}\varphi}(\xi)=\langle\xi\rangle^{-2}\widehat\varphi(\xi)$, where $\langle\xi\rangle=\sqrt{1+|\xi|^2}$, for all $\varphi\in L^2$ Apr 28 at 7:44
• @LL3.14 Yes I think that's the right idea, but we need to use the Sobolev inequality on a bounded domain in order to not have an infinite sum. Does this work? By the Sobolev embedding theorem, we have $\|{f}\|_{L^4(\square_\alpha)}\leq C(|\square|)\|{f}\|_{H^s(\square_\alpha)}$ if $s=d/4$. We will need that $s\leq1$, so $d\leq 4$. Apr 28 at 7:58

By defining $$\psi = (1-\Delta)\varphi$$, 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 $$d=2$$, the Sobolev inequality does not work anymore from $$H^1$$ to $$L^\infty$$, but still works to any $$L^p$$ with $$p\in[2,\infty)$$.