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I am studying p-admissible functions. I am using the book of heinonen (nonlinear potential theory of degenerate elliptic equations). I am searching for a good proof of the result:

Suppose that w is a p-admissible weight and q > p. Then w is q-admissible" .

The proof of heinonen book is much difficult. Someone can give me a suggestion for a reference for a proof of this result?

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Post you doubts in the demonstration and maybe we can help you. – Tomás Feb 19 '13 at 13:58
up vote 0 down vote accepted

I am gonna try to explain all the demonstration and with this, I am helping you and improving my skills.

We have al locally integrable function, nonegative function in $\mathbb{R}^n$ and we define the Radon measure $$\mu(E)=\int_E w(x)dx$$

$w$ is said to be $p$-admissible if it satisfies I,II,III and IV below.

$\bf{\mbox{I - (Doubling Property)}}$ $0<w<\infty$ a.e. in $\mathbb{R}^n$ and there exist a constant $C_I>0$ such that $$\mu(2B)\leq C_I\mu(B)$$

This is trivial.

$\bf{\mbox{II - (The Gradient is Well Definite)}}$ If $D$ is an open set and $\phi_i\in C^\infty(D)$ is a sequence of functions such that $\int_D |\phi_i|^qd\mu\rightarrow 0$ and $\int_D |\nabla\phi_i-v|^qd\mu\rightarrow 0$, then $v=0$

If $G$ is a open bounded set, then $L^q(G)$ is continuously embedded in $L^p(G)$. This implies that $$\int_G |\phi_i|^pd\mu\rightarrow 0\ \mbox{and}\ \int_G |\nabla\phi_i-v|^pd\mu\rightarrow 0 $$

Because $w$ is $p$-admissible, you conclude that $v=0$ in $G$. Now, write $D=\cup G_i$ where $G_i$ is open, bounded and $\overline{G}\subset D$.

$\bf{III - \mbox{(Weighted Sobolev Inequality)}}$ There are constants $\varkappa>1$ and $C_{III}>0$ such that $$\tag{1} \Big(\frac{1}{\mu(B)}\int_B |\phi|^{\varkappa q}d\mu\Big)^{\frac{1}{\varkappa q}}\leq C_{III}r\Big(\frac{1}{\mu(B)}\int_B |\nabla\phi|^q d\mu\Big)^{\frac{1}{q}}$$

whenever $B=B(x_0,r)$ is a ball in $\mathbb{R}^n$ and $\phi\in C_0^{\infty}(B)$.

Let $\psi\in C_0^\infty(B)$ and $\phi=\psi^+$. As Pavel alread showed to you here, if $s=\frac{q}{p}$, then $\phi^s\in W^{1,p}_0(B)$. Because $C_0^\infty(B)$ is dense in $W^{1,p}_0(B)$, you can find a sequence $\phi_n\in C_0^\infty(B)$ such that $\|\phi_n-\phi^s\|_{1,p}\rightarrow 0$. Note that the sequence $\phi_n$ satisfies (1) for each $n$. Now you can apply Lebesgue Theorem to conclude that $\phi^s$ also satisfies (1). From this we get

\begin{eqnarray} \Big(\frac{1}{\mu(B)}\int_B (\phi^s)^{\varkappa p}d\mu\Big)^{\frac{1}{\varkappa p}}y &=& C_{III}sr\Big(\frac{1}{\mu(B)}\int_B |\nabla\phi|^p\phi^{(s-1)p} d\mu\Big)^{\frac{1}{p}} \nonumber \\ &\leq& \tag{2} C_{III}sr\Big(\frac{1}{\mu(B)}\int_B |\nabla\phi|^{sp} d\mu\Big)^{\frac{1}{sp}}\Big(\frac{1}{\mu(B)}\int_B \phi^{sp} d\mu\Big)^{\frac{s-1}{sp}} \end{eqnarray}

where in the last inequality, we have used Holder inequality (note that $\frac{1}{s}+\frac{s-1}{s}=1$).

Moreover, you can use Holder in the functions $\phi^{sp}$ and $1$ (use the exponents $\varkappa$ and $\varkappa'$, where $\frac{1}{\varkappa}+\frac{1}{\varkappa'}=1$), to conlcude that $$\tag{3} \Big(\frac{1}{\mu(B)}\int_B \phi^{sp}d\mu\Big)^{\frac{1}{p}}\leq\Big(\frac{1}{\mu(B)}\int_B \phi^{\varkappa sp}d\mu\Big)^{\frac{1}{\varkappa p}}$$

Combining (2) and (3) and using the facto that $s=\frac{q}{p}$, you conclude that $$\Big(\frac{1}{\mu(B)}\int_B \phi^{\varkappa q}d\mu\Big)^{\frac{1}{\varkappa q}}\leq c\frac{q}{p}r\Big(\frac{1}{\mu(B)}\int_B |\nabla\phi|^{q}d\mu\Big)^{\frac{1}{q}}$$

where $c$ is an positive constant not depending on $B$ nor $\phi$. By doing the same calculation, you can verify that the last inequality also is valid for the negative part of $\psi$: $\max\{0,-\psi\}$. Now you use the fact $|\psi|=\max\{0,\psi\}+\max\{0,-\psi\}$, to conclude the demonstration.

$\bf{IV - (\mbox{Weighted Poincaré Inequality})}$ There is a constant $C_{IV}>0$ such that $$\int_B |\phi-\phi_B|^q d\mu\leq C_{IV}r^p\int_B |\nabla\phi|^q d\mu$$ whenever $B=B(x_0,r)$ is a ball in $\mathbb{R}^n$ and $\phi\in C^\infty(B)$ is bounded. Here $$\phi_B=\frac{1}{\mu(B)}\int_B\phi d\mu$$

Let $\phi\in C^\infty(B)$ and $\gamma\in\mathbb{R}$. Note that

\begin{eqnarray} \mu(B)|\gamma-\phi_B|^q &=& \mu(B)\Big|\frac{1}{\mu(B)}\int_B (\gamma-\phi)d\mu\Big|^q \nonumber \\ &\leq& \mu(B)\Big(\frac{1}{\mu(B)}\int_B|\gamma-\phi|d\mu\Big)^q \nonumber \\ &\le& \mu(B)\frac{\mu(B)^{\frac{q}{q'}}}{\mu(B)^q}\int_B |\phi-\gamma|^qd\mu \\ &=& \int_B |\phi-\gamma|^qd\mu \end{eqnarray}

I have used Holder with $\frac{1}{q}+\frac{1}{q'}=1$. It follows that

\begin{eqnarray} \int_B|\phi-\phi_B|^qd\mu &=& \int_B|\phi-\gamma+\gamma-\phi_B|^q \nonumber \\ &\leq& 2^{q-1}\int_B |\phi-\gamma|^q+2^{q-1}\int_B |\gamma-\phi_B|^{q} \nonumber \\ &\le& 2^q\int_B |\phi-\gamma|^qd\mu \end{eqnarray}

Because of the last inequality, it is sufficient to prove that there exists constants $\gamma$ and $C$ such that $$\int_B |\phi-\gamma|^qd\mu\leq Cr^q\int_B |\nabla\phi|^qd\mu$$

Let $s=\frac{q}{p}$ and write $$v=\max\{\phi-\gamma,0\}^s-\max\{-(\phi-\gamma),0\}^s$$

where $\gamma$ is chosen so that $\int_B vd\mu=0$ (why such $\gamma$ exist?). Note that $v\in W^{1,p}$ (why?) and again, by using a argument of density, we can conclude that $$\int_B |v|^pd\mu\leq C_{IV}r^p\int_B |\nabla v|^pd\mu$$

Note that $|v|=\max\{\phi-\gamma,0\}^s+\max\{-(\phi-\gamma),0\}^s=|\phi-\gamma|^s$ and hence, $|\nabla v|=s|\nabla\phi||v|^{\frac{s-1}{s}}$. From here I think you can finish, it is the same history: you substitue $|\nabla v|$ in the last inequality, use Holder, ...

Notes: Try to adpat the demonstration of Lemma 1.11, to construct the convergent sequence that we have used in the proof of III and IV.

Also, I have used Lemma 1.19, so take a look on thi lemma. If you have any doubt, please post here.

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Thank you ! there are many days that i am working to understand the proof of heinonen. your proof will help me a lot ! Thank you ! (my english is terrible, sorry ) – math student Feb 21 '13 at 15:10
No Problem, my english is bad too (Sou brasileiro tambem!). – Tomás Feb 21 '13 at 15:46
tentei chegar na formula $|∇v|=s|∇ϕ||v|^{(s-1)/s}$ , mas nao consegui... tem como vc me dar um dica ? – math student Feb 21 '13 at 20:16
You can consider two cases: $\phi-\gamma>0$ and $\phi-\gamma<0$. In the first case you have that $v=(\phi-\gamma)^s$ which implies $\nabla v=s(\phi-\gamma)^{s-1}\nabla\phi$ and hence $|\nabla v|=s|\phi-\gamma|^{s-1}|\nabla\phi|$. The second case is analogous. Now use the characterization of $|v|$ that I gave you: $|v|=|\phi-\gamma|^s$ which implies $|v|^{\frac{s-1}{s}}=|\phi-\gamma|^{s-1}$ – Tomás Feb 21 '13 at 22:16

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