Let $\displaystyle \Omega\subset\mathbb{R}^n$ be a bounded open set. Prove the inclusion $\displaystyle W^{1,n}(\Omega)\subset L^{p}(\Omega)$ for any $\displaystyle p\in[1,\infty)$.

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- Sep 9th 2012, 06:55 AMkierkegaardInclusion in Sobolev spaces
Let $\displaystyle \Omega\subset\mathbb{R}^n$ be a bounded open set. Prove the inclusion $\displaystyle W^{1,n}(\Omega)\subset L^{p}(\Omega)$ for any $\displaystyle p\in[1,\infty)$.

- Sep 13th 2012, 03:29 AMjohnsomeoneRe: Inclusion in Sobolev spaces
It's been a while, so this is just to see if I've unpacked the problem correctly.

$\displaystyle u \in W^{1,n}(\Omega) \Rightarrow u \in L^{n}(\Omega)$ and $\displaystyle D_iu \in L^{n}(\Omega)$ for $\displaystyle i \in \{1, 2, ... N \}$, where $\displaystyle D_i$ is the weak partial derivative in the $\displaystyle x_i$ coordinate.

Want to show that if for some $\displaystyle u$ that's true for some $\displaystyle n$ (I assume $\displaystyle n \ge 1$ ??), then $\displaystyle u \in L^{p}(\Omega) \ \forall p \in [1,\infty)$.

Since $\displaystyle |\Omega| < \infty$, it follows that $\displaystyle ||u||_p \rightarrow ||u||_{\infty}$.

It would suffice to prove the inclusion if could show $\displaystyle ||u||_{\infty} < \infty$ (although that might not be the case - it's just an initial thought as to how to begin.)

What $\displaystyle D_iu$ exists means is that there are functions $\displaystyle D_iu \ni \int_{\Omega} u\frac{\partial \phi}{\partial x_i} = - \int_{\Omega} \phi D_iu \ \ \ \forall \phi \in C_{c}^{\infty}(\Omega)$.

Anyway, I just wanted to see if I have the setup correct. I don't know the common techniques for Sobolev spaces, but what I'll be thinking about is: Can show $\displaystyle u$ is bounded a.e.? What about showing just $\displaystyle ||u||_1 < \infty$? What if $\displaystyle u \in C^1(\Omega)$, since it's "close" to that? What about approximating $\displaystyle u$ by smooth functions? Can show $\displaystyle ||u||_p < \infty$ on compact subsets that expand to $\displaystyle \Omega$ in a controllable way so that that the limit is also finite? Surely $\displaystyle \Omega$ boounded is important - how should that be exploited? I dunno - it's just initial random thoughts.

I'd be interested to see how it's done, since I imagine it's done using common techniques for this topic. If I get anywhere, I'll post it.

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One more observation. The condition that $\displaystyle D_iu \in L^{n}(\Omega)$ for $\displaystyle i \in \{1, 2, ... , N \}$, the Sobolev-ness, is vital, since without it, the claim fails.

Let $\displaystyle \Omega = ( 0, 1 ), u(x) = \frac{1}{\sqrt{x}}$. Then $\displaystyle u \in C^{\infty}(\Omega) \cap L^1(\Omega)$, but $\displaystyle u \notin L^2(\Omega)$.

$\displaystyle u$ failed to be in $\displaystyle W^{1,1}(\Omega)$ because $\displaystyle u'(x) = \frac{-1}{2\sqrt{x^3}} \notin L^1(\Omega)$.

In the same vein, suppose $\displaystyle f \in W^{1,n}(0,1)$ is of the form $\displaystyle f(x) = x^{\alpha}$. Then:

$\displaystyle f'(x) = \alpha x^{\alpha - 1} \in L^n(0,1) \Rightarrow n(\alpha - 1)>-1 \Rightarrow \alpha >1 - 1/n \geq 0$

$\displaystyle \Rightarrow p\alpha \geq 0 \ \forall \ p \geq 1 \Rightarrow f(x)^p = x^{p\alpha} \in L^1(0,1) \ \forall \ p \geq 1$.

Thus $\displaystyle f(x) = x^{\alpha} \in W^{1,n}(0,1) \Rightarrow f \in L^p(0,1) \ \forall \ p \geq 1$. It also implies that$\displaystyle ||f||_{\infty} < \infty$.

Perhaps an the idea for the proof is found by: $\displaystyle D_1u \in L^n$ somehow gets you, using $\displaystyle u \in L^n$, that $\displaystyle u \in L^m$ for some m > n. And then repeat that reasoning with this bigger m, "staircasing" your way up to higher and higher values so that can eventually conclude $\displaystyle u \in L^p \ \forall p$?

(If this approach even worked, I wouldn't expect $\displaystyle D_1u$ to "staircase" up - it would remain stuck in $\displaystyle L^n$ - since otherwise the whole of $\displaystyle W^{1,n}$ would staircase up to be a subset of $\displaystyle W^{1,p} \ \forall p$, unless of course that's actually true, or true for bounded domains (I simply don't know -but I suspect it's not so).)