Gradient inequality in Morrey’s and Gagliardo-Nirenberg inequality proofs

calculusfunctional-analysisintegrationreal-analysissobolev-spaces

I'm approaching for the first time functional analysis and in general "advanced math". In the book "Lecture Notes on Functional Analysis" by A. Bressan, in the Sobolev space chapter, I get stuck in two inequalities involving $\nabla u$. $$$$ 1) In the proof of Morrey's inequality: be $u \in C^{1}(\mathbb{R}^{n})\cap W^{1,p}(\mathbb{R}^{n})$; in the proof, the following set of coordinates is defined: $(r,\xi)=(r,\xi_{2}…\xi_{n}) \in \left[0,\rho \right] \times B_{1}$, with $B_{\rho}:=\left( x=(x_{1},x_{2},…,x_{n});x_{1}=\rho, \ \sum_{i=2}^{n} \ x^2_{i}\leq \rho^{2}\right) $; to this system of coordinates, the point $x(r, \xi)=(r,r \xi)$ is associated (x belongs to a cone), moreover define $U(r,\xi)=u(r,r\xi)$. Here I got stuck (citing the text): since $|\xi|\leq 1$ the directional derivative of $u$ in the direction of the vector $(1,\xi_{2},\xi_{3},…,\xi_{n})$ is estimated by $$\left| \frac{\partial }{\partial r} U(r,\xi)\right|=\left| u_{x_{1}}+\sum_{i=2}^{n}\xi_{i}u_{x_{i}} \right| \leq 2|\nabla u(r,\xi)|.$$
$$ $$ 2) In the proof of Gagliardo-Nirenberg inequality: be $f \in C^{1}_{c}\left( \mathbb{R}^{n} \right)$ , I got stuck in the following inequality: $$ \prod _{i=1}^{n} \left( \int _{-\infty}^{\infty}…\int _{-\infty}^{\infty} \left | D _{ x _{i}} dx _{1} dx _{2}… dx _{n} \right | \right) ^{1/(n-1)} \leq \left ( \int _{\mathbb{R}^{n}}^{} |\nabla f|dx \right )^{n/(n-1)} $$ $$$$ It seems to me that I miss something becouse in the book it appears that such inequalities involving $\nabla u$ and $ \nabla f $ are so straightforward; I probalby lack the knowledge of some identity or inequality, perhaps from calculus. Could anyone give me some hints? Thank you for your kindness, I apologize if it might be trivial, but I’m a rookie!

Best Answer

The first one is Cauchy-Schwarz: $$ \big|\sum_{i=2}^n\xi_iu_{x_i}\big| \leq \big(\sum_{i=2}^n\xi_i^2\big)^{\frac12} \big(\sum_{i=2}^n|u_{x_i}|^2\big)^{\frac12} \leq|\xi||\nabla u| \leq|\nabla u|. $$

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