Discussion Board Lecture 04
Two small comments on lecture 4
| [#16] | ||
|---|---|---|
Dear ISemTeam, first, I want to point out two typos: In the third line of the proof of Theorem 4.16 one should replace $ \|g\|_2\|v\|_{2,1} $ by $ \|f\|_2\|v\|_{2,1} $ and in the title of the second reference one should read Analyse fonctionnelle in lieu of Analyse fonctionelle. Second, here is an alternative access to Lemma 4.5, which avoids the use of Exercise 4.5 and leads perhaps a little bit more directly to the conclusion $ f=0 $. Instead of referring to Lemma 4.3 it is based on Corollary 4.4: We put $ \Omega_n:=\{x\in\Omega:\ |x|<n\} $ and $ A_m:=\{x\in\Omega:\ |f(x)|\leq m\} $ for $ n,m\in\mathbb N $. Then $ f_{n,m}:=(f\chi_{A_m})|_{\Omega_n}\in L^2(\Omega_n) $ and, by hypothesis, the bounded linear functional $ (\,\cdot\,|f_{n,m})_{L^2(\Omega_n)} $ vanishes on the dense subspace $ C_c^\infty(\Omega_n) $ of $ L^2(\Omega_n) $. Hence, $ (\,\cdot\,|f_{n,m})_{L^2(\Omega_n)}=0 $ and, in particular, $ 0=(f_{n,m}|f_{n,m})=\|f_{n,m}\|^2_{L^2(\Omega_n)} $, i.e., $ f_{n,m}=0 $ a.e. on $ \Omega_n $ for all $ m\in\mathbb N $. Since the countable union of zero sets is itself a zero set, we deduce $ f|_{\Omega_n}=0 $ a.e. on $ \Omega_n $, which finally yields $ f=0 $ as asserted. Best wishes, Heiko
|
||
Dear Heiko, I'm amazed by people who find such typos even in the References! Thanks a lot for pointing out those two things. Your proof of Lemma 4.5 looks nice, but I don't understand yet why the functional $ (\cdot|f_{n,m})_{L_2(\Omega_n)} $ vanishes on $ C_{\rm c}^\infty $. Doesn't the cut-off at $ m $ cause problems? Best wishes, Hendrik
| ||
Dear Hendrik, it is in fact Peer Kunstmann who amazed you :-). He has found that misprint and I have only communicated it. Concerning your question, I've recognized that I used a result from measure theory which does not emerge immediately from the content of the lecture: By the lecture we know that we may find a sequence $ (\varphi_k)_k $ in $ C_c^\infty(\Omega_n) $ converging to $ \chi_{A_m\cap\Omega_n} $ in $ L_1(\Omega_n) $, but in fact one can even achieve that additionally $ 0\leq \varphi_k\leq 1 $ and that $ (\varphi_k)_k $ converges also pointwise almost everywhere to $ \chi_{A_m\cap\Omega_n} $. Then one obtains $ 0=\lim_{k\to\infty}\int_\Omega f\varphi\varphi_k=\int_{\Omega_n} f_{n,m}\varphi $ for each $ \varphi\in C_c^\infty(\Omega_n) $ by dominated convergence. In view of this I apologize for suggesting this alternative, which is not so helpful. Best wishes, Heiko
| ||
Dear Heiko, OK, then I can say that Peer doesn't cease to amaze me :-) Thanks for the explanation of your alternative proof. One minor nitpick: I have no problems with achieving that $ (\varphi_k)_k $ is a.e. convergent; this is well-known measure theory. I also know how to achieve $ 0\leqslant\varphi_k\leqslant1 $, but I can't say I could do it with just measure theory. I'd use composition with a suitable $ C^\infty $-version of a cut-off function. That said, I find your suggestion very helpful! Let me propose a variant of your argument: For $ n\in\mathbb N $ we put $ A_n := \{x\in \Omega{;}\; |f(x)|\leqslant n,\ |x|\leqslant n\} $. Then $ \mathbf{1}_{A_n}\mkern1mu\bar f \in L_1(\Omega) $, so by Corollary 4.4 there exists a sequence $ (\varphi_k)_k $ in $ C_{\rm c}^\infty(\Omega) $ such that $ \varphi_k \to \mathbf{1}_{A_n}\mkern1mu\bar f $ in $ L_1(\Omega) $ as $ k\to\infty $. One can even achieve that additionally $ |\varphi_k|\leqslant n+1 $ and that $ (\varphi_k)_k $ is a.e. convergent. Then one obtains $ 0 = \lim_{k\to\infty}\int_\Omega f\varphi_k = \int_\Omega|f|^2\mathbf{1}_{A_n} $, i.e., $ f=0 $ a.e. on $ A_n $. Since the countable union of zero sets is itself a zero set, we deduce that $ f=0 $ a.e. I wouldn't say that this variant is really simpler than the one in the lecture, but I like it, too. What do you think? (By the way, I wrote "$ \leqslant n+1 $" above since this is most easily achieved by composition with a $ C^\infty $-function $ h\colon\mathbb R\to\mathbb R $ with $ h(x)=x $ for all $ |x|\leqslant n $ and $ |h(x)|\leqslant n+1 $ for all $ x\in\mathbb R $.) Best wishes, Hendrik
| ||
Dear Hendrik, yes, I agree, that this variant is not really simpler - and for this reason I've classified it "not so helpful" in hindsight :-) - and thanks for your variant of the suggested argument. Best wishes, Heiko P.S.: "I can't say I could do it with just measure theory" Indeed, I should have been preciser saying something like "measure and integration theory in $ \mathbb R^d $"...
| ||
Comments
| [#17] | ||
|---|---|---|
Dear isem team, firstly, on behalf of the Darmstadt team I'd like to thank you for a nice lecture. The only real remark we have concerns Remark 4.10 (a), as we don't understand how to generalize the proof of Theorem 4.9 to the case where $ a=\infty $ or $ b=\infty $. We would be grateful if you could provide further explanations. The following remarks are minor points: In your definition of locally integrable functions it may be helpful to mention that your definition is equivalent to "integrable on every compact set", at least at our university this definition seems to be more prominent. Next, in the proof of Theorem 4.8 there was a little confusion about which norm the sum of the $ L^2 $'s is endowed with. Finally, when you define the support of a locally integrable function you refer to an "already defined support for continuous functions" but we couldn't find this previous definition. At last we would like to mention two typos: In the first line of Chapter 4.2 there is a "the this", which probably should be just "this". And in the proof of Lemma 4.17 there should be a closed bracket at the very end. Regards Alex
|
||
Hi Alex, in the proof of Theorem 4.9 it seems that the norm of the inclusion is related to the length of the interval (which would be bad news in the case of an unbounded interval), but in fact it is not: Given $ x \in (-\infty,b] $ you can choose an interval $ J_x \subseteq (-\infty, b] $of length $ 1 $ containing $ x $. Then of course $ f \in H^1(J_x) $ and the proof of Theorem 4.9 gives $ \|f\|_{C(J_x)} \leq 2 \|f\|_{H^1(J_x)}\leq 2 \|f\|_{H^1(-\infty,b)} $. As this estimate is valid for every $ x \in [\infty,b] $ you get $ \|f\|_{C((-\infty,b])} \leq 2 \|f\|_{H^1(-\infty,b)} $. Regards, Moritz
| ||
Dear Alex, the support of a continuous function was defined in Exercise 1.2 (a) in the first lecture. Best wishes,
| ||
Dear Alex, Moritz has already explained the main point. (Thanks!) I want to add why one also can conclude that $ f(x)\to 0 $ as $ x\to-\infty $. This is because $ \|f\|_{C(J_x)} \leq 2 \|f\|_{H^1(J_x)} $, and $ \|f\|_{H^1(J_x)}\to 0 $ as $ x\to-\infty $. Concerning `locally': Strictly speaking, a property is said to hold locally if every point has a neighbourhood where this property holds. This is the general convention. So, if a function is locally integrable, then also on compact sets, by a standard proof. The converse holds because $ \Omega $ is locally compact. Concerning the orthogonal direct sum: We had hoped that it is accepted as standard that the norm in an orthogonal direct sum should be $ \|(f^0,\ldots,f^n)\|=(\sum_0^n\|f^j\|^2)^{1/2} $. And thanks for the typos! Best wishes, Jürgen
| ||
Dear all, thank you very much for your explanations and clarifications. Regards, Alex
| ||
