Discussion Board Lecture 12
Question about remark 12.7 c)
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Dear ISEM Team, Congratulations for these very interesting lectures. I have a little question about remark 12.7 c); the hypothesys are that $ a $ is a closed sectorial form in $ H $ with domain $ V $, with $ H, V $ Hilbert spaces: in the sequel it is written that this implies that $ a $ defines a norm (instead of a semi-norm) on $ V $. Why? Best regards, Giorgio Menegatti
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Dear Giorgio, the semi-norm $ \|\cdot\|_a $ defined on $ {\rm dom}(a) $ (see Section 12.1) is the norm mentioned on line 4 of the remark you refer to, and it is a norm because the norm of $ H $ is included. You may have overlooked this. Best wishes, Jürgen
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Remarks and questions from Darmstadt
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Dear ISem Team, we did enjoy reading through lecture 12 and want to thank you again for the lecture. Two questions came up. Firstly, on page 133, remark 12.1, should there not be a stronger condition on $ a $? For instance positivity or accretivity, just so that the spuareroot in (12.2) is well-defined. Moreover on page 143 in exercise 12.3 (a) the density is according to which space? We thought of $ H^1(\Omega ) $. And finally we would use this discusion board to ask a question that came up while talking about Hausdorff measures. Is there a bounded open set $ \Omega \subset \mathbb{R}^n $, such that the boundary has Hausdorff dimension smaller that $ n-1 $? Thanks for any ideas. Best regards, the Darmstadt Team.
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Dear Marc, concerning your final question on the Hausdorff measure of the boundary of bounded open sets the answer is "no": If $ \Omega \subseteq \mathbb{R}^n $ is bounded, then $ \partial \Omega $ has Hausdorff dimension at least $ n-1 $. This is for the following reason. Write $ \mathbb{R}^n = \mathbb{R} \times \mathbb{R}^{n-1} $ and consider the projection $ P $ onto $ \mathbb{R}^{n-1} $. You can check by the very definition of Hausdorff measure/dimension that applying $ P $ will not increase the Hausdorff dimension, that is, the Hausdorff dimension of $ \partial \Omega \subseteq \mathbb{R}^n $ is greater or equal to the Hausdorff dimension of $ P(\partial \Omega) \subseteq \mathbb{R}^{n-1} $. The upshot is that $ P(\partial \Omega) \supseteq P(\Omega) $ holds (this is because for every $ x \in \Omega $ the line passing through $ x $ in direction $ (1,0) $ exits the bounded set $ \Omega $ through a boundary point. The set $ P(\Omega) $ is open in $ \mathbb{R}^{n-1} $, hence has full Hausdorff dimension $ n-1 $. This implies that $ P(\partial \Omega) $ has full Hausdorff dimension $ n-1 $ as well and the claim follows. What do you think? Best, Moritz
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Dear Marc, oh yes, you're right – we forgot to assume that $ a $ is positive in Remark 12.1. Thanks for pointing this out! As for the Hausdorff dimension of the boundary of a bounded open set in $ \mathbb R^n $, Moritz gave a very good answer. What he wrote for the projection $ P $ is true in a more general setting: If $ P $ is a contraction of $ \mathbb R^n $ (i.e. Lipschitz with constant $ 1 $), then for the $ d $-dimensional Hausdorff measure of a Borel set $ A\subseteq\mathbb R^n $ one has $ \sigma_d(P(A))\le\sigma_d(A) $. Best wishes, Hendrik
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Dear Hendrik, of course you are right, what I was using has nothing to do with projections. Something else I just realized later (completely off-topic, though) is that I was answering the question of my own team, whose meeting I missed on monday -- welcome to the age of internet :-D Best, Moritz
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Dear Darmstadt Team, you may be interested in the fact that the $ n-1 $-dimensional Hausdorff measure of the essential boundary (also called perimeter) of a set with finite and not 0 measure, not only is not 0, but it is greater or equal to the perimeter of a ball with the same measure: this is the isoperimetric inequality, and it was proved by De Giorgi in 1958. Best regards, Giorgio Menegatti
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Some questions and comments on Lecture 12
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Dear ISemTeam, I have some questions and comments concerning Lecture 12. 1. On page 136 the inequality (*) $ \operatorname{Re}b(u_n)=\operatorname{Re}(b(u_n,u_n-u_k))+\operatorname{Re}(b(u_n,u_k))\leq \|u_n\|_b\cdot\|u_n-u_k\|_b+|(Bu_n|u_k)_H| $ is displayed and justified by reference to (12.2). Could you make your argument more explicit, please? I guess that you want to estimate separately the absolute values of the two summands on the left hand side and that (12.2) is to be used in order to estimate the absolute value of the first one. Is that right? However, if I use (12.2), then I only come up with $ \operatorname{Re}b(u_n)=\operatorname{Re}(b(u_n))=\operatorname{Re}(b(u_n,u_n-u_k))+\operatorname{Re}(b(u_n,u_k)) \leq|\operatorname{Re}(b(u_n,u_n-u_k))|+|\operatorname{Re}(b(u_n,u_k))| $ $ \leq|b(u_n,u_n-u_k)|+|b(u_n,u_k)| \leq (1+c)\|u_n\|_b\cdot\|u_n-u_k\|_b+|(Bu_n|u_k)_H| $, which is clearly sufficient for the following reasoning, but not the asserted inequality. Alternatively, one may apply directly Proposition 5.2 and use that the form $ (\operatorname{Re}b) $ is positive, as $ B $ is sectorial, thus accretive; then the above factor $ 1+c $ does indeed not appear at all, but here one does not need (12.2) and the estimate for the second term is different (nevertheless this inequality is also sufficient for the subsequent argumentation): $ \operatorname{Re}b(u_n)=|(\operatorname{Re}b)(u_n)| \leq|(\operatorname{Re}b)(u_n,u_n-u_k)|+|(\operatorname{Re}b)(u_n,u_k)| $ $ \leq\left(\operatorname{Re}b(u_n)\operatorname{Re}b(u_n-u_k)\right)^{\frac{1}{2}}+\frac{1}{2}|(Bu_n|u_k)_H+\overline{(Bu_k|u_n)_H}| $ $ \leq\|u_n\|_b\cdot\|u_n-u_k\|_b+\frac{1}{2}\left(|(Bu_n|u_k)_H+(B^*u_n|u_k)_H|\right) $, where I used that $ B^* $ is an operator since $ B $ is densely defined (Theorem 6.3 (a)).
$ \operatorname{Re}a(u)+\omega\|u\|_H^2\geq\alpha\|j(u)\|_V^2 $ should be replaced by $ \operatorname{Re}a(u)+\omega\|j(u)\|_H^2\geq\alpha\|u\|_V^2 $.
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Dear Heiko, could you please put some line break into your 2 longest formulas? The text is so wide that it doesn't fit onto my screen :-( Thanks a lot, Hendrik P.S.: One thing in advance: $ \operatorname{Re}a(u,v) $ always means $ \operatorname{Re}(a(u,v)) $; otherwise we'd write $ (\operatorname{Re}a)(u,v) $.
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Dear Hendrik, I've edited my first posting in order to put some line breaks (sorry for the inconvenience with those long lines!) and to incorporate the hint of your postscript (but my question seems to remain open). Best wishes, Heiko
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Dear Heiko, thanks, now it's much easier to read your post. It appears that you're right on all counts (which doesn't surprise me :-)). Thanks a lot for pointing out those mistakes / gaps! 1. I did notice quite some time ago that a factor $ (1+c) $ is missing in the proof of Theorem 12.5, but somehow I managed to convince myself that I was mistaken and that it's also true without that factor, because of the $ \operatorname{Re} $. You're right, IF everything is read with the symmetric form $ \operatorname{Re}b $, then the factor $ (1+c) $ is not needed. However, the argument that you gave how to proceed in this version does not work: one could well have $ \operatorname{dom}(B) \cap \operatorname{dom}(B^*) = \varnothing $, and then $ B^*u_n $ makes no sense. In short: Use (12.2), and insert the missing factor $ (1+c) $. 2. Yes, of course! 3. You're right, the argument as it is now isn't complete. However, I'd prefer to say that obviously $ \tilde a $ is sectorial since $ a $ is sectorial. Then Corollary 5.11 only tells us that $ \tilde A $ is quasi-m-sectorial, but using Proposition 5.5 we see that the operator is even m-sectorial. The whole situation in Lecture 5 is somewhat unfortunate in the following respect: we're lacking a concise result telling us that a $ j $-elliptic sectorial form is associated with an m-sectorial operator. Theorem 5.9 almost says this, but there we assumed coerciveness instead of $ j $-ellipticity, which wasn't defined yet at that point. (I just hope that I didn't produce any new blunders above.) Best wishes, Hendrik P.S.: Note that Jürgen wrote a new reply to your post concerning Lecture 10.
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Dear Hendrik, thank you for your explanations, and you are perfectly right: the second argument is expedient only under the additional assumption that $ \operatorname{dom}(B)\cap\operatorname{dom}(B^*) $ is dense in $ V $ (what a silly mistake!). Best wishes, Heiko
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Some missprints
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Dear ISem team, I found some missprints: p. 138, middle, displayed line: $ k\to\infty $, not $ n\to\infty $ (2 times) Best wishes,
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Dear Johannes, thanks a lot for pointing out those misprints! On p. 141 there are indeed several integrals without a "dx", but for those near the bottom that was by intention. For the other three I guess it wasn't ... Best wishes, Hendrik
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Dear Johannes, thanks for the $ dx $'s. They are indeed missing: in contexts where also $ d\sigma $ occurs we prefer to write all the $ dx $'s. However, sorry, I found only 6! Am I missing one of the missing ones? But I want to point out another slip. The Robin Laplacian should carry the other sign. As it is defined here, it is not consistent with Lecture 7. Best wishes, Jürgen O.K. The missing $ dx $ was found!
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