Let be a bounded open set. Prove the inclusion for any .
It's been a while, so this is just to see if I've unpacked the problem correctly.
and for , where is the weak partial derivative in the coordinate.
Want to show that if for some that's true for some (I assume ??), then .
Since , it follows that .
It would suffice to prove the inclusion if could show (although that might not be the case - it's just an initial thought as to how to begin.)
What exists means is that there are functions .
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 is bounded a.e.? What about showing just ? What if , since it's "close" to that? What about approximating by smooth functions? Can show on compact subsets that expand to in a controllable way so that that the limit is also finite? Surely 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 for , the Sobolev-ness, is vital, since without it, the claim fails.
Let . Then , but .
failed to be in because .
In the same vein, suppose is of the form . Then:
.
Thus . It also implies that .
Perhaps an the idea for the proof is found by: somehow gets you, using , that 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 ?
(If this approach even worked, I wouldn't expect to "staircase" up - it would remain stuck in - since otherwise the whole of would staircase up to be a subset of , unless of course that's actually true, or true for bounded domains (I simply don't know -but I suspect it's not so).)