Lemma about the weighted interpolation inequality

Question

In this article Interpolation inequalities with weights Chang Shou Lin the following lemma is stated and proved.

Lemma: Suppose $\dfrac{1}{p}+\dfrac{\alpha}{n} > 0$, then there exists a constant $C$ such that

$$|||x|^{\alpha}u||_p \leq C |||x|^{\alpha+1}Du||_p.$$

the proof consists only of the direct calculation \begin{align} |||x|^{\alpha}u||_p^p& = \int|x|^{p\alpha}|u|^p dx\\ & \leq C \int|x|^{p\alpha+1}|u|^{p-1}|Du| dx\\ & \leq C \left( \int|x|^{p\alpha}|u|^p dx \right)^{1-1/p} \left( \int|x|^{p(\alpha+1)}|Du|^p dx \right)^{1/p} \end{align} I'm almost certain that in the first inequality, we just used integration by parts. Hence, it is necessary to assume that the function u has compact support (the lemma does not make it clear what the function's class is), I got $$-(p\alpha+1)\int |x|^{p\alpha} \dfrac{\vec{x}}{|x|} u^pdx = p \int |x|^{p\alpha+1}u^{p-1}Dudx $$ I guess I must be missing something because I'm not using the hypothesis $\dfrac{1}{p}+\dfrac{\alpha}{n} > 0$. Or maybe that's not how it's done. Does anyone have any suggestions? Also, I'm thinking that the proper space in which this inequality holds is the space of C^1 functions with compact support, is that the best space? The last inequality is Holder's inequality.

Answer 1

$\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}$In view of conditions (0.1) and (0.4) in the linked paper, $p\in[1,\infty)$ and $u\in C_0^\infty(\R^n)$. To prove the lemma in question (Lemma 2.2 in that paper), it suffices that $u\in C_0^1(\R^n)$.

Indeed, \begin{equation*} \|\,|x|^\al u\|_p^p=I:=\int_{\R^n}dx\,|x|^{p\al} |u(x)|^p. \tag{5}\label{5} \end{equation*} In view of the condition $u\in C_0^1(\R^n)$,
\begin{equation*} -|u(x)|^p=\int_1^\infty dt\,pu(tx)^{[p-1]}(Du)(tx)\cdot x, \end{equation*} where $z^{[p-1]}:=|z|^{p-2}z$ for real $z\ne0$, with $0^{[p-1]}:=0$, $(Du)(y)$ is the gradient of $u$ at $y$, and $\cdot$ denotes the dot product. So,
\begin{equation*} \begin{aligned} I&\le\int_{\R^n}dx\,|x|^{p\al+1} \int_1^\infty dt\,p|u(tx)|^{p-1}|Du|(tx) \\ &=C\int_{\R^n}dy\,|y|^{p\al+1} \,p|u(y)|^{p-1}|Du|(y), \end{aligned} \end{equation*} where \begin{equation*} C:=\int_1^\infty \frac{dt}{t^{n+p\al+1}}=\frac1{n+p\al}, \end{equation*} given the condition \begin{equation*} \frac1p+\frac\al n>0, \tag{10}\label{10} \end{equation*} which can be rewritten as $n+p\al>0$.

So, using Jensen's inequality as in your post, we get \begin{equation*} I\le C I^{1-1/p}\|\,|x|^{\al+1}Du\|_p. \tag{20}\label{20} \end{equation*} Using now conditions $u\in C_0^1(\R^n)$ and \eqref{10} again, we see that \begin{equation*} I\le C_2\int_{\R^n}dx\,|x|^{p\al}1(|x|\le C_2)=C_3\int_0^{C_2} dr\, r^{p\al+n-1}<\infty, \end{equation*} where $C_2,C_3$ are positive real numbers, possibly depending on $u$. Thus, by \eqref{5} and \eqref{20}, \begin{equation*} \|\,|x|^\al u\|_p\le C\| |x|^{\al+1}Du\|_p. \quad\Box \end{equation*}

Answer 2

Take polar coordinates, and you find $$\int |x|^{p\alpha} |u|^p dx = \int_{\mathbb{S}^{n-1}} \int_0^\infty r^{p\alpha + n - 1} |u|^p ~dr ~dA$$ This you rewrite as $$ C \int_{\mathbb{S}^{n-1}} \int_0^\infty |u|^p ~d(r^{p\alpha + n}) ~dA$$ and perform the inner integral by parts. In order for the integration by parts to not generate any boundary terms at $r = 0$, you need $r^{p\alpha + n}|_{r = 0} = 0$ which requires $p\alpha + n > 0$.