Case n=2 in boundedness of subelliptic operator (Gilbarg-Trudinger Theorem 8.15)

Question

Theorem 8.15. along with relevant inequalities below.

It have two questions about the proof. First, how do we know that $H(w) \in W^{1,2}_0(\Omega)$? Wouldn't we need some bounds on $\beta\ge 1$?

Other than this, I followed the proof for the case where $n \ge 3$. However, if $n=2$, I get a little lost. If $n>2$, then the Sobolev inequality states that $W_0^{1,2}(\Omega)$ is embedded in $L^{\frac{2n}{n-2}}(\Omega)$. However, for $n=2$ this result is not valid. Therefore, I don't see how to obtain $$ \lVert H(w) \rVert_{\frac{2\cdot \hat{2}}{\hat{2}-2}} \le C\left(\int_\Omega \overline{b}\left(w H^\prime(w)\right)^2\,\mathrm{d}x\right)^{1/2}. $$

Equation 8.3

\begin{equation} Lu=D_i(a^{ij}(x)D_ju+b^i(x)u)+c^i(x)D_iu+d(x)u \end{equation}.

Equation 8.29 \begin{equation} \begin{aligned} A^i(x,z,p) &= a^{ij}(x)p_j + b^i(x)z - f^i(x),\\ B(x,z,p) &= c^i(x)p_i + d(x)z - g(x) \end{aligned} \end{equation} Equation 8.30 For $v\ge 0, v\in C_0^1(\Omega)$

\begin{equation} \int_{\Omega}\left(D_ivA^i(x,u,Du)-vB(x,u,Du)\right)dx \le(\ge,=)0 \end{equation}

Equation 8.32

\begin{equation} \bar z=|z|+k,\qquad \bar b=\lambda^{-2}(|b|^2+|c|^2+k^{-2}|f|^2)+\lambda^{-1}(|d|+k^{-1}|g|) \end{equation}

Equation 8.33

\begin{align} p_iA^i(x,z,p) & \ge \frac{\lambda}{2}(|p|^2-2\bar b\bar z^2) \\ | \bar zB(x,z,p) | &\le \frac{\lambda}{2}\left( \epsilon|p|^2+\frac{\bar b}{\epsilon}\bar z^2\right) \end{align}

Any help will be greatly appreciated

Answer 1

First Question: No, we don't need an upper bound on $\beta$. Indeed, we can explicitly write that $H: [k,\infty) \to \mathbb R$ is given by $$H(z) = \begin{cases} z^\beta - k^\beta, &\text{if } z\in [k,N) \\ \beta N^{\beta-1}(z-N)+(N^\beta-k^\beta), &\text{if } z \geqslant N. \end{cases} $$ We are able to calculate the value of $H$ in the region $z\geqslant N$ explicitly since we know $H$ is linear in that region and we know that $H$ is continuously differentiable across $z=N$. Hence, $$H'(z) = \begin{cases} \beta z^{\beta-1}, &\text{if } z\in [k,N) \\ \beta N^{\beta-1}, &\text{if } z \geqslant N, \end{cases} $$ so $\vert H'(z)\vert \leqslant \beta{N^{\beta-1}}, $ that is, $H'$ is bounded. Since, $w\in W^{1,2}(\Omega)$, we conclude that $H(w) \in W^{1,2}(\Omega)$ by the chain rule in Sobolev spaces. To see that $H(w) \in W^{1,2}_0(\Omega)$, we use that $w=k$ on $\partial \Omega$ in the trace-sense and that $H(k)=0$, so $H(w)=0$ on $\partial \Omega$ in the trace-sense.

Second question: Let $n=2$ and $\hat n \in (2,q)$ be arbitrary. Let $\alpha:=\frac{2 \hat n}{\hat n -2}$ and note that $\alpha>2$. Now set $p = \frac{2\alpha}{\alpha+2}$ and observe that $p \in (1,2)$ and $$ \frac{2\hat n }{\hat n -2} = \frac{2p}{2-p}.$$ Hence, by the Sobolev inequality, \begin{align*} \| H(w)\|_{L^{\frac{2\hat n}{2-\hat n}}(\Omega)} &=\| H(w)\|_{L^{\frac{2p}{p-2}}(\Omega)} \leqslant C_\ast(\hat n) \|D (H(w))\|_{L^p(\Omega)}. \end{align*} Then, using that $p< 2$, by Hölder's inequality, $$\|D (H(w))\|_{L^p(\Omega)} \leqslant \vert \Omega \vert^{\frac 1 p - \frac 1 2}\|D (H(w))\|_{L^2(\Omega)}. $$ Thus, the inequality holds with $C(\hat n , \vert \Omega \vert)=C_\ast(\hat n ) \vert \Omega \vert^{\frac 1 p - \frac 1 2}$ for any $\hat n >2$.