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

Question

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!

Answer 1

1. It is as timur said for the first one - \begin{align} \left| \frac{\partial }{\partial r} U\right|\overset{\substack{chain\\rule\\{\phantom{x}}\\}}=\left| u_{x_{1}}+\sum_{i=2}^{n}\xi_{i}u_{x_{i}} \right| \overset{\triangle ineq.}\le |u_{x_{1}}|+\sum_{i=2}^{n}|\xi_{i}||u_{x_{i}}| \overset{C-S}\le |u_{x_{1}}|+\sqrt{\sum_{i=2}^{n}|\xi_{i}|^2 } \sqrt{\sum_{i=2}^{n}|u_{x_{i}}|^2 } \end{align} Now observe that \begin{align}|u_{x_1}| &\le \sqrt{\sum_{k=1}^n |u_{x_k}|^2} = |\nabla u|,\tag{GradientBound}\label{X}\\ \sqrt{\sum_{i=2}^{n}|\xi_{i}|^2 } &\le 1, \\ \sqrt{\sum_{k=2}^n |u_{x_k}|^2} &\le \sqrt{\sum_{k=\color{red}1}^n |u_{x_k}|^2}=|\nabla u|,\end{align} which gives $ \left| \frac{\partial }{\partial r} U\right| \le 2|\nabla u|$ as claimed.

2. I can confidently interpret $\prod _{i=1}^{n} \left( \int _{-\infty}^{\infty}...\int _{-\infty}^{\infty} \left | D _{ x _{i} dx _{1} dx _{2}... dx _{n} } \right | \right) ^{1/(n-1)} \leq \left ( \int _{\mathbb{R}^{n}}^{} |\nabla f|dx \right )^{n/(n-1)}$ as this part of the proof of the Gagliardo-Nirenberg-Sobolev inequality: $$\prod _{i=1}^{n} \left( \int _{-\infty}^{\infty}\dots\int _{-\infty}^{\infty} | D_{ x _{i}}f| dx _{1} dx _{2}\dots dx _{n} \right) ^{1/(n-1)} \leq \left ( \int _{\mathbb{R}^{n}}^{} |\nabla f|dx \right )^{n/(n-1)}$$ Indeed, this is what page 171 of Bressan's book says. This is also the proof in Evans.

3. To prove this, note that there is a product of $n$ terms on the left hand side, and there is an $n$ power on the left. This hints that there may be an estimate that does not depend on which of the $n$ terms you are estimating. That is, it is enough to show that for each $i=1,2,\dots,n$,

$$ \left( \int _{-\infty}^{\infty}\dots\int _{-\infty}^{\infty} | D_{ x _{i}}f| dx _{1} dx _{2}\dots dx _{n} \right)^{1/(n-1) } \le \left(\int _{-\infty}^{\infty}\dots\int _{-\infty}^{\infty} |\nabla f| dx_1dx_2\dots dx_n\right)^{1/(n-1)}$$ and this directly follows from the line marked $\eqref{X}$ above.

Answer 2

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