# Nash Inequality

As an exercise from my book I wanted to prove:

For $$n>2$$ and for any $$u \in W_0^{1,2}\left(\mathbb{R}^n\right) \cap L^1\left(\mathbb{R}^n\right)$$, $$\int_{\mathbb{R}^n}|\nabla u|^2 d x \geq c\left(\int_{\mathbb{R}^n} u^2 d x\right)^{\frac{n+2}{n}}\left(\int_{\mathbb{R}^n}|u| d x\right)^{-\frac{4}{n}},$$ where $$c=c(n)>0$$.

I solved a similar problem and used an appropriate Hölder inequality and the Sobolev inequality in the form $$\left(\int_{\mathbb{R}^n}|u|^{\frac{2 n}{n-2}} d x\right)^{\frac{n-2}{n}} \leq C \int_{\mathbb{R}^n}|\nabla u|^2 d x$$ and I think it works here too, I just cannot find the fitting exponents. Any hints on how to tackle this one?

• Apply Holder to $\int u^2$ and then Sobolev to one of the resulting factors. Oct 28, 2023 at 21:06
• @Deane I did the same for the Poincaré Inequality, the problem is how to get that term with $-4/n$. Oct 28, 2023 at 21:09
• Move it to the other side. Oct 28, 2023 at 21:27
• @Deane I made an edit and tried to use Hölder Oct 30, 2023 at 21:43
• You applied Holder to $(u)(1)$. That won't work. In particular, the constant will not depend on the volume of $\Omega$. Oct 30, 2023 at 22:34

## 1 Answer

By Hölder's inequality $$\int |u|^2 = \int |u|^{\frac{4}{n+2} + \frac{2n}{n-2}\frac{n-2}{n+2}} \\ \leq \left(\int |u|\right)^\frac{4}{n+2} \left(\int |u|^\frac{2n}{n-2}\right)^\frac{n-2}{n+2}$$ By Sobolev's inequality, which reads (notice a mistake in your question about the exponent) $$\left(\int |u|^\frac{2n}{n-2}\right)^\frac{n-2}{n} \leq C_n \int |\nabla u|^2$$ this implies $$\int |u|^2 \leq C_n^\frac{n}{n+2}\left(\int |u|\right)^\frac{4}{n+2} \left(\int |\nabla u|^2\right)^\frac{n}{n+2}$$ or equivalently $$\left(\int |u|^2\right)^\frac{n+2}{n} \leq C_n\left(\int |u|\right)^\frac{4}{n} \int |\nabla u|^2.$$