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I am reading the following paper (preprint here) and the author starts by stating the Sobolev inequality on the Sphere $\mathbb{S}^d$ $$\frac{p-2}{d}\int |\nabla u|^2 + \int |u|^2 \geq \left(\int |u|^p\right)^{2/p} \tag 1$$ where the integral is taken over the Sphere and $2<p\leq \frac{2d}{d-2}.$ The author mentions that this inequality can be derived by considering the Sobolev inequality on $\mathbb{R}^d$ with optimal constant, which is \begin{align} \int |\nabla u|^2 \geq S \int |u|^p \tag2 \end{align}

where $S$ is the best constant. I am not sure how to deduce $(1)$ from $(2)$ using Stereographic projection. Any comments/remarks will be much appreciated.

Edit: I came across the following preprint: https://arxiv.org/pdf/1010.5821.pdf which explains how to deduce the above inequality. Essentially, the computations boil down to deducing (A.3) from (A.2) [see page 9].

Given $f\in L^2(\mathbb{S}^n)$ we define $F\in L^2(\mathbb{R}^n)$ such that $$F(x) = \sqrt{J_S(x)} f(S(x))$$ where $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $S(x)=\left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right)$ and the jacobian of the map $J_{S}(x) = \left(\frac{2}{1+|x|^2}\right)^{n}.$ Then I want to show that, $$\int_{\mathbb{R}^n}|\nabla F|^2 dx = \int_{\mathbb{S}^n} |\nabla f|^2 + \frac{n(n-2)}{4}|f|^2 d\omega.$$

I tried the compute the derivatives using the representation of $F$ but the computations do not seem to simplify.

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    $\begingroup$ This is explained in detail in the book "Analysis" of Lieb and Loss, section 4.4. $\endgroup$ Commented Nov 15, 2021 at 16:07
  • $\begingroup$ @GiuseppeNegro thanks for sharing this reference! $\endgroup$
    – Student
    Commented Nov 15, 2021 at 16:12
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    $\begingroup$ One difficulty is computing the gradient of a function $f$ on the sphere. My suggestion is to extend the function to a function $\tilde{f}: \mathbb{R}^n\backslash\{0\} \rightarrow \mathbb{R}$ homogeneous of degree $0$. In other words, let $$ \tilde{f}(x) = f\left(\frac{x}{|x|}\right). $$ Then the gradient of $\tilde{f}$ is equal to the spherical gradient of $f$, because $x\cdot \nabla \tilde{f} = 0$. $\endgroup$
    – Deane Yang
    Commented Nov 28, 2021 at 23:42

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