I should first mention this: in my studies of Sobolev spaces I have completed all the questions of chapter 9 from Folland's real analysis with the help of this site and this is my last one, which is also very interesting, but seemingly difficult so I am posting here in the hopes of getting some help, but I should first mention some notation conventions used here.

$ \mathcal{S} $ is the Schwartz class of functions and $ \mathcal{S}' $ is the class of distributions ("generalized functions") on $ \mathcal{S} $

$ \partial ^ {\alpha} $ is the multi-index distribution derivative on $ L^2 $

$ \Lambda _ s f = { [(1+|\xi|^2)^{\frac{s}{2}} \widetilde{f} ]^{\vee} } $ is a continuous linear operator on the Schwartz distributions.

The Sobolev space $ H_s = \{ f \in S' | \Lambda_s f \in L^2 \} $

Now define the following differential operator $ D^{\alpha} = (2i\pi)^{-\alpha}\partial^{\alpha} $ along with its symbol $ P(D) = \sum_{|\alpha|<m} (2\pi i)^{|\alpha|}a_{\alpha} D^{\alpha} $, this of course has the property that the Fourier transform of $ P(D)[f] $ is equal to $ P\widehat{(f)} $

And now the elliptic regularity theorem as I know it:

Suppose that L is a constant-coefficient elliptic differential operator of order m and $ \Omega $ is open in $ R^n $ and $ u \in D'(\Omega) $. If $ Lu \in H_{s}^{loc}(\Omega) $ for some $ s \in R $, then $ u \in H_{s+m}^{loc}(\Omega) $. Also, if $ Lu \in C^{\infty}(\Omega) $ then $ u \in C^{\infty}(\Omega) $.

I can honestly say this theorem makes sense to me. Now for the exercise at hand:

The heat operator $ \partial_t - \Delta $ is not elliptic but a weakened version of the elliptic regularity theorem does hold for it. Here we are workign on $ R^{n+1} $ with coordinates $ (x,t) $ and dual coordinates $ (\xi,\tau) $, and $ \partial_t - \Delta = P(D) $ where $ P(\xi,\tau) = 2i\pi\tau + 4 {\pi}^2 |\xi|^2 $.

a. There exist $ C,R > 0 $ such that $ |\xi| |(\xi,\tau)|^{\frac{1}{2}} \leq C |P(\xi,\tau)| $ for $ |(\xi,\tau)| > R $ (We are to try to consider the region $ |\tau| \leq |\xi|^2 $ and its complement separately).

b. If $ f \in H_s $ and $ (\partial_t - \Delta)f \in H_s $ then we are to show $ f \in H_{s+1} $ and $ \partial _ {x_i} f \in H_{s+(1/2)} $ for $ 1 \leq i \leq n $

c. If $ \zeta \in C_C^{\infty}(R^{n+1}) $, then we are to show: $ [\partial_t - \Delta,\zeta]f = (\partial_t \zeta - \Delta \zeta)f - 2 \sum{(\partial_{x_i}\zeta)(\partial_{x_i}f)} $

d. We are to show that if $ \Omega \in R^{n+1} $ is open and $ u \in \mathcal{D}'(\Omega) $ and if $ (\partial_t-\Delta)u \in H_{s}^{loc}(\Omega) $ then $ u \in H_{s+1}^{loc}(\Omega) $. We are given this hint for part d:

enter image description here

Here is the proof of theorem 9.26 the elliptic regularity theorem: enter image description here

My problem is this is fairly advanced stuff and I cannot seem to find it in me to write this down rigorously despite being able to see the geometric intuition behind it, but still this is hard analysis and I am a novice... I am positng this here following the successes I had and knowing this is the last question I have before I can truly say I completed this chapter's exercises. Thanks in advance to all helpers.

********** Progressed I think I am only truly stuck on parts b and d where for part d I cannot seem to imitate the proof referenced.

  • 2
    $\begingroup$ As a humble physicist, wish I could only decipher the above .. $\endgroup$ – Han de Bruijn Nov 1 '15 at 19:24
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    $\begingroup$ @HandeBruijn what do you mean exactly? $\endgroup$ – kroner Nov 1 '15 at 19:32
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    $\begingroup$ If I'm understanding correctly $b$ is false in general: If $u$ is a constant function, then $(\partial_t-\Delta)u=0\in L^2(\mathbb{R}^{n+1})$, but $u\notin L^2(\mathbb{R}^{n+1})$. $\endgroup$ – Jose27 Nov 15 '15 at 2:46
  • $\begingroup$ @Jose27 No but the constant function does not satisfy the second assumption despite it being zero under heat operator $\endgroup$ – kroner Nov 15 '15 at 3:32
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    $\begingroup$ Right, thanks. For some reason I missed the hypothesis $f\in H_s$ completely. $\endgroup$ – Jose27 Nov 15 '15 at 4:22

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