Sobolev Inequality for $u \in W_{0}^{1,p}(\mathbb{R}^{n})$.

cauchy-sequencesfunctional-analysislp-spacespartial differential equationssobolev-spaces

For $1\leq p<n$ and $$p^{*}:=\frac{np}{n-p},$$ one can prove that there exists a constant $c>0$ such that $$\|u\|_{L^{p^{*}}(\mathbb{R}^{n})}\leq c\|\nabla u\|_{L^{p}(\mathbb{R}^{n})}$$ for all $u\in C_{0}^{\infty}(\mathbb{R}^{n})$. This is known as the Sobolev-Gagliardo-Nirenberg inequality.

I was trying to prove that this must also hold for all $u\in W_{0}^{1,p}(\mathbb{R}^{n})$. The space $W_{0}^{1,p}(\mathbb{R}^{n})$ is the $\|\cdot\|_{W^{1,p}(\mathbb{R}^{n})}$-closure of $C_{0}^{\infty}(\mathbb{R}^{n})$ in the Sobolev space $W^{1,p}(\mathbb{R}^{n})$, where $\|u\|_{W^{1,p}(\mathbb{R}^{n})}^{p}:=\|u\|_{L^{p}(\mathbb{R}^{n})}^{p}+\|\nabla u\|_{L^{p}(\mathbb{R}^{n})}^{p}$.

To do so, I tried to use a density argument: If $u\in W_{0}^{1,p}(\mathbb{R}^{n})$, then there exists a sequence $(u_{k})$ in $C_{0}^{\infty}(\mathbb{R}^{n})$ such that $\|u_{k}-u\|_{W^{1,p}(\mathbb{R}^{n})}\to0$. Using the Sobolev-Gagliardo-Nirenberg inequality and the definition of $\|\cdot\|_{W^{1,p}(\mathbb{R}^{n})}$, we see that $$\|u_{k}-u_{l}\|_{L^{p^{*}}(\mathbb{R}^{n})}\leq c\|\nabla(u_{k}-u_{l})\|_{L^{p}(\mathbb{R}^{n})}\leq c\|u_{k}-u_{l}\|_{W^{1,p}(\mathbb{R}^{n})}$$ for all $k,l\geq1$. So $(u_{k})$ is Cauchy in the Banach space $L^{p^{*}}(\mathbb{R}^{n})$ and therefore convergent. Thus there is a $\tilde{u}\in L^{p^{*}}(\mathbb{R}^{n})$ such that $\|u_{k}-\tilde{u}\|_{L^{p^{*}}(\mathbb{R}^{n})}\to0$.

I don't know how to finish the argument from here. Can we for example conclude that $\tilde{u}=u$ a.e.? Any help would be greatly appreciated! Thanks in advance.

Best Answer

Since $u_k\to u$ in $W^{1,p}\subset L^1_{loc}$ ($L^1$ when restricted to compact subsets of $\mathbb R^n$) and $u_k\to \tilde u$ in $L^{p^*} \subset L^1_{loc}$, it follows that $u=\tilde u$. So $u=\tilde u\in L^{p^*}\cap W^{1,p}$.

Furthermore, if $(X,\|\cdot\|_X)$ is any normed space, then by rearranging the triangle inequality $\|u+v\|_X \le \|u\|_X + \|v\|_X$ we obtain $$ |\|u\|_X-\|v\|_X| \le \|u-v\|_X$$ Applying this to our Banach spaces, $$\|u_k\|_{L^{p^*}} \to \|u\|_{L^{p^*}} ,\quad \|\nabla u_k \|_{L^p} \to \|\nabla u\|_{L^p}$$

Thus we may pass to the limit in the norm inequality.

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