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If we let $\Omega\subset\mathbb{R}^d$ with $d=1,2,3$ and define $\mathcal{H}^1(\Omega)=(w\in L_2(\Omega): \frac{\partial w}{\partial x_i}\in L_2(\Omega), i=1,...,d)$. My tutor has repeated several times:

  1. If $d=1$ then $\mathcal{H}^1(\Omega)\subset\mathcal{C}^0(\Omega)$.
  2. If $d=2$ then $\mathcal{H}^2(\Omega)\subset\mathcal{C}^0(\Omega)$ but $\mathcal{H}^1(\Omega)\not\subset\mathcal{C}^0(\Omega)$.
  3. If $d=3$ then $\mathcal{H}^3(\Omega)\subset\mathcal{C}^0(\Omega)$ but $\mathcal{H}^2(\Omega)\not\subset\mathcal{C}^0(\Omega)$.

I was interested in trying to show these relationships. Does anyone know any references that would be useful.

Thanks in advance.

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2  
The classical reference is of course Adams and Fournier: Sobolev spaces. That book is not easy going, though. –  Harald Hanche-Olsen May 21 '10 at 11:44
    
See also mathoverflow.net/questions/17736/… –  Willie Wong May 21 '10 at 17:54

4 Answers 4

up vote 7 down vote accepted

I understand from your post that you'd like to show those facts by yourself first, and not necessarily to approach the whole theory now (I like your approach). Trivial hint: start with smooth functions with compact support in $\Omega$, and try to bound their $L^\infty$ norm in terms of the $H^d$ norm. Also, I suggest that you try building counter-examples by yourself for the case of non-inclusions. Reference: Brezis' book of Functional Analysis may give you nice hints.

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In the Hilbert space setting the easiest way to see the whole thing is to fo on the Fourier side and see what you can get from Cauchy-Schwartz (just localize first by multiplying by a smooth cutoff). The second part of Claim 3 is wrong by the way... –  fedja May 22 '10 at 1:52

If you just want the answer, then not surprisingly you can find it at: http://en.wikipedia.org/wiki/Sobolev_inequality

If you want a careful introduction to and derivation of the Hilbert space case, see: "Seminar on the Atiyah-Singer Index Theorem" (Princeton Univ. Press)

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By now, I can't remember precisely where the best places to learn this is. Here are some rather vague suggestions:

  1. I don't know if this stuff is in any of Nirenberg's writings, but if it is, it's sure to be a clear and easy approach.

  2. Look in books about nonlinear elliptic PDE's by, say, Craig Evans, Gilbarg and Trudinger, or Thierry Aubin.

  3. Ideally, there should be a proof that involves integration over cubes. Differential geometers such as Aubin tend to prove such results by integrating over balls because that's what generalizes more easily to Riemannian manifolds. That works fine but the formulas are messier than for a cube. In the end, after you get the idea of what's going on, just write out your own proof.

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I'll give you a hint for the first one $d=1$. Consider first the case that your function $f \in H^1([0,1])$ was smooth. Then we could say $f(x) - f(y) = \int_{x}^y f'(s)ds$. Apply Cauchy Schwartz now and you'll be able to see immediately that $f$ is $1/2$ Hölder continuous.

For higher dimensions you actually proceed similarly but you need to use the co-area formula.

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Cheers! :) The first one $d=1$ is simple doing it this way. Now off to find out about the co-area formula... –  alext87 Sep 6 '10 at 19:19
    
Also, for the $d=2$ case, consider the function $u(x) = \log |x|$. Then $|Du(x)| = \frac{1}{|x|}$ and so $\int |Du(x)|^2 = 2\pi \int_0^r r^2/r$ which is finite obviously. So $\log|x|$ is in $H_0^1(\Omega)$ but not continuous. A similar example can be made for $d=3$ with $1/|x|^{\alpha}$ for an appropriate choice of $\alpha > 0$. –  Dorian Sep 6 '10 at 19:46

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