Using Nash inequality to derive an inequality (from proof in paper)

We work on a domain $\Omega \subseteq \mathbb{R}^N$ with the Dirichlet Laplacian. Let $\lVert \cdot \rVert_p$ denote the $L^p$ norm.

I am trying to understand why the following inequality is true: for $u$ such that the following norms are sensible, $$\lVert u\rVert_{p}^p\lVert (-\Delta)^{\sigma/ 4}|u|^{\frac{p+m-1}{2}}\rVert_2^2 \geq C\lVert u\rVert^{2p+m-1}_{\frac{N(2p+m-1)}{2N-\sigma}}$$ from the proof of Theorem 8.2 in this work http://arxiv.org/pdf/1104.0306.pdf.

To do this, the authors use the Nash--Nagliardo--Nirenberg inequality

N-G-N inequality: $$\lVert v \rVert_{r_2}^{\alpha + 1} \leq C\lVert (-\Delta)^{\gamma / 2}v \rVert_r \lVert v \rVert_p^\alpha$$ where $p \geq 1$, $r > 1$, $0 < \gamma < \min(N,2)$ and $r_2 := \frac{N(rp+r-p)}{r(N-\gamma)}$ and $\alpha = \frac{p(r-1)}{r}$

with the choice $r=p+m-1$.

I don't get this at all. Surely we need to pick $r=2$ in order to get the fractional term right.

• Do you see that one can use the NGN inequality with $r=2$ and suitable $r_{2}$, $q$ (substituting $q$ for $p$ to avoid double usage), and $\alpha$? Or are you asking for details of that argument too? – Matt Rosenzweig Jan 20 '16 at 0:45
• @MattRosenzweig With $r=2$ and suitable $r_2$ and $q$ I don't get the desired inequality, so i don't see it, so detail of the calculation would be good. Of course, as I quoted, the authors say we should use a different choice of $r$, and I don't get that either. So details would be appreciated – 25Chars Jan 20 '16 at 12:43

Rewrite the inequality that you want to prove as $$\|u\|_{p}^{p/2}\|(-\Delta)^{\sigma/4}|u|^{\frac{p+m-1}{2}}\|_{2}\geq C\|u\|_{\frac{N(2p+m-1)}{2N-\sigma}}^{\frac{2p+m-1}{2}}$$ Set $v:=|u|^{\frac{p+m-1}{2}}$. The LHS above becomes $$\|v\|_{q}^{\alpha}\|(-\Delta)^{\sigma/4}v\|_{2}$$ with $q=\frac{2p}{p+m-1}$ and $\alpha=\frac{p}{p+m-1}$. Apply the NGN inequality with $r=2$ and $q$ and $\gamma=\sigma/2$. In general $q$ need not satisfy the conditions of the NGN inequality, but if you look at the bottom of pg. 25, they say that $p_{k}$ satisfies $$\min\{1,m\}\leq\frac{p_{k}}{p_{k}+m-1}\leq\max\{1,m\},$$ so I think we're ok in this context. You can check that with $r$, $q$, and $\alpha$ above, we have that $$r_{2}=\frac{N(q+2)}{2N-\sigma}$$ so we have $$\|v\|_{q}^{\alpha}\|(-\Delta)^{\sigma/4}v\|_{2}\geq\|v\|_{r_{2}}^{\alpha+1}$$ But \begin{align*} |v|^{r_{2}}=|u|^{(\frac{p+m-1}{2})(\frac{N(q+2)}{2N-\sigma})}=|u|^{\frac{N(2p+m-1)}{2N-\sigma}} \end{align*} and \begin{align*} \frac{\alpha+1}{r_{2}}=\frac{2p+m-1}{p+m-1}\cdot\frac{2N-\sigma}{N(\frac{2p}{p+m-1}+2)}=\frac{(2N-\sigma)(2p+m-1)}{2N(2p+m-1)} \end{align*} So $$\|v\|_{r_{2}}^{\alpha+1}=\|u\|_{\frac{N(2p+m-1)}{2N-\sigma}}^{\frac{2p+m-1}{2}}$$