How to prove the homogeneous Sobolev spaces $\dot{W}^{1,p}$ on cubes are equivalent to the Sobolev spaces

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Let $1\leqslant p<\infty$, $Q\subset \mathbb{R}^N$ a cube. Define the homogeneous Sobolev space: $$ \dot{W}^{1,p}(Q)=\{u\in L^1_{\mathrm{loc}}(Q): D^\alpha u\in L^p(Q), |\alpha|=1\}.$$ Prove that $\forall u\in \dot{W}^{1,p}(Q)$, $u\in W^{1,p}(Q)$.

I know if $u\in \dot{W}^{1,p}(Q)$, then $u\in L^p_{\mathrm{loc}}(Q)$, hence $\forall \tilde{Q}\subset\subset Q$, we get $u\in W^{1,p}(\tilde{Q})$, where "$\subset\subset$" means $\tilde{Q}$ is open, the closure of $\tilde{Q}$ is compact and contained in $Q$. I don't know what to do next. This is the variant of Exercise 13.35 come from A First Course in Sobolev Spaces, 2nd written by Giovanni Leoni. This exercise is given behind Poincare's inequality for rectangles, but I don't know how to use this proposition. enter image description here

And I also want to know whether this claim is true or not for $p=\infty$.

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Let $T\Subset S\Subset R$. Then $u\in W^{1,p}(S)$. By the fundamental theorem of calculus, for all $x\in S$ and $y\in T$, \begin{align*} |u(x)-u(y)| & \leq|u(x)-u(x_{1},\ldots,x_{N-1},y_{N})|\\ & \quad+\cdots+|u(x_{1},y_{2},\ldots,y_{N})-u(y)|\\ & \leq\sum_{i=1}^{N}\int_{0}^{a_{i}}|\partial_{i}u(x_{1},\ldots ,x_{i-1},t,y_{i+1},\ldots,y_{N})|\,dt. \end{align*} Raising both sides to power $p$, by H"{o}lder's inequality, and the convexity of the function $g(t)=|t|^{p}$, \begin{align*} |u & (x)-u(y)|^{p}\leq\biggl(\sum_{i=1}^{N}\Bigl(\int_{0}^{a_{i}} |\partial_{i}u(x_{1},\ldots,x_{i-1},t,y_{i+1},\ldots,y_{N})|^{p} dt\Bigr)^{\frac{1}{p}}\biggr)^{p}\\ & \leq N^{p-1}\sum_{i=1}^{N}\int_{0}^{a_{i}}|\partial_{i}u(x_{1} ,\ldots,x_{i-1},t,y_{i+1},\ldots,y_{N})|^{p}dt. \end{align*} Hence, again by H"{o}lder's inequality and the fact that $u(x)-u_{T} =(u(x)-u)_{T}$, \begin{align*} \int_{S} & |u(x)-u_{T}|^{p}\,dx\leq\frac{1}{(\mathcal{L}^{N}(T))^{p}}\int _{S}\Bigl(\int_{T}|u(x)-u(y)|\,dy\Bigr)^{p}dx\\ & \leq\frac{1}{(\mathcal{L}^{N}(T))^{p-p/p^{\prime}}}\int_{S}\int _{T}|u(x)-u(y)|^{p}dydx\\ & \leq\frac{N^{p-1}}{\mathcal{L}^{N}(T)}\sum_{i=1}^{N}\int_{S}\int_{T} \int_{0}^{a_{i}}|\partial_{i}u(x_{1},\ldots,x_{i-1},t,y_{i+1},\ldots ,y_{N})|^{p}dtdydx\\ & \leq\frac{N^{p-1}\mathcal{L}^{N}(R)}{\mathcal{L}^{N}(T)}\sum_{i=1}^{N} a_{i}\int_{R}|\partial_{i}u(z)|^{p}dz\leq\frac{N^{p-1}\mathcal{L}^{N} (R)}{\mathcal{L}^{N}(T)}\sum_{i=1}^{N}\int_{R}|\partial_{i}u(z)|^{p}dz, \end{align*} where we have used the facts that $p-p/p^{\prime}=1$, $\max\{a_{1} ,\ldots,a_{N}\}=1$, and Fubini's theorem.

If you now take an increasing sequence $S_{n}$ containing $T$ and whose union is $R$, replacing $S$ with $S_{n}$ and using the Lebesgue monotone convergence theorem on the left-hand side gives $$ \int_{R}|u(x)-u_{T}|^{p}\,dx\leq\frac{N^{p-1}\mathcal{L}^{N}(R)} {\mathcal{L}^{N}(T)}\sum_{i=1}^{N}\int_{R}|\partial_{i}u(z)|^{p}dz, $$