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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?

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

1 Answer 1

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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 $$ \left(\int |u|^2\right)^\frac{n+2}{n} \leq C_n\left(\int |u|\right)^\frac{4}{n} \int |\nabla u|^2. $$

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