1

We are in $\mathbb R^d$. Let $\frac{1}{p}+\frac{1}{q}=1$ and $q=\frac{2d}{d-2}$, $d\geq 3$. I suppose $f\in L^q(\mathbb R^d)$, $\nabla f\in L^2(\mathbb R^d)$ and $g\in L^p(\mathbb R^d)\cap L^2(\mathbb R^d)$. I denote $$\left<f,g\right>=\int_{\mathbb R^d}fg,$$ and $G$ is the radial solution of $\Delta u=0$, i.e. $G(x,y)=K|x-y|^{d-2}$ where $K$ is a constant. So, I proved that $$|\left<f,g\right>|^2\leq \|\nabla f\|_{L^2(\mathbb R^d)}^2\left<g,G*g\right>.$$

Moreover, I know that $$\|f\|_{L^q(\mathbb R^d)}=\sup\{|\left<f,g\right>|\mid \|g\|_{L^p(\mathbb R^d)}\leq 1\},$$ and thus

$$\|f\|_{L^q(\mathbb R^d)}^2=\sup\{|\left<f,g\right>|^2\mid \|g\|_{L^p(\mathbb R^d)}\leq 1\}\leq \|\nabla f\|_{L^2(\mathbb R^d)}^2\sup\{\left<g,G*g\right>\mid \|g\|_{L^p}\leq 1\}.$$

Now, my book (Analysis second edition by Elliott Lieb and Michael Loss page 203) says : By taking $g=f^{q-1}\in L^p(\mathbb R^d)$ we finally obtain $$\|f\|_{L^q(\mathbb R^d)}^{2q}\leq \|\nabla f\|_{L^2(\mathbb R^d)}^2\left<f^{q-1},G*f^{q-1}\right>,$$ but I really don't know how to get it.

Any idea ?

user349449
  • 1,577
  • For $g=f^{q-1}$ you have $|\langle f,g\rangle|^2=|f|_{L^q}^{2q}$. – Jeff Jan 07 '18 at 17:21
  • @Jeff: Really ? How do you get that ? to me $$|\left<f,f^{q-1}\right>|=\left|\int f^q\right|^2\leq \left(\int|f|^q\right)^2=|f|_{L^q}^{2q}.$$ But we don't have equality. – user349449 Jan 07 '18 at 18:17
  • Good point. For $q$ even, they are the same. For odd $q$ you could replace $f$ by $|f|$ in the original inequality. – Jeff Jan 07 '18 at 18:29
  • @Jeff : But $q$ is neither odd nor even, it's even not an integer a priori. By the way, if $q$ would be odd, why could you replace $f$ by $|f|$ ? it's not the same... – user349449 Jan 07 '18 at 20:05
  • Well, $q=2d/(d-2)$ where $d$ is an integer, but you are right this is not always an integer. The point I was making is that $|f|f^{q-1}=|f|^q$ when $q$ is odd (and $|\nabla f|{L^2} = |\nabla |f||{L^2}$). – Jeff Jan 07 '18 at 20:42
  • Notice there is an argument on page 204 showing that you can reduce to the case where $f$ is nonnegative; this clears up any issues you have I think. – Jeff Jan 07 '18 at 20:56

0 Answers0