Discussion Board Lecture 10
Some remarks on Lecture 10
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Dear ISemTeam, here are some remarks concerning Lecture 10: 1) I think instead of Theorem 10.1 one should call Corollary 10.3 "the three lines theorem" since three lines actually appear only there :-). If one wishes to give Theorem 10.1 a name, I think it would be more convenient to speak of a special case of the Phragmén-Lindelöf theorem. 2) There is a small gap concerning definitions: Throughout the lecture $ \operatorname{sgn} u $ is used for complex-valued functions $ u $, but it seems that $ \operatorname{sgn} u $ has been defined so far only for real-valued functions (in Lemma 9.9). 3) On page 112 almost at the end of the proof of Lemma 10.7, it is claimed that there exists $ A\in\mathcal{A}_c $ with $ \mu(A)>0 $ and $ A\subseteq\lbrack|u|>c\rbrack $ provided that $ \lbrack|u|>c\rbrack $ is not a null set. Could you please give more details how to derive this assertion? In general, this assertion is wrong: Consider, e.g., $ \Omega=\lbrack 0,1\rbrack $ with $ \mathcal A $ the Borel $ \sigma $-algebra and $ \mu $ the (Borel-)Lebesgue measure and let $ \mathcal{A}_c $ be the ring generated by the left open subintervals of $ \lbrack 0,1\rbrack $. Furthermore, let $ C\subseteq\lbrack 0,1\rbrack $ be a Cantor set of positive measure. Then there is no element in $ \mathcal{A}_c $ of positive measure contained in $ C $. By the way, here is a slight variant to complete the proof: Suppose to the contrary that $ \lbrack|u|>c\rbrack $ is not a null set. Since the countable union of null sets is itself a null set there exists an $ n\in\mathbb N $ such that $ B:=\lbrack|u|>c\rbrack\cap A_n $ has positive $ \mu $-measure (notice that for each $ m\in\mathbb N $ we have $ \lbrack|u|>c\rbrack\cap A_m=\lbrack |u|\chi_{A_m}>c\rbrack $, hence these sets are indeed elements of $ \mathcal A $ since $ |u|\chi_{A_m} $ is measurable thanks to $ |u|\chi_{A_m}\in L_1(\mu) $; alternatively, one may use the observation that the sequence $ v_k:=u\cdot\chi_{\bigcup_{n=1}^kA_n} $ of measurable (cf. the preceding argument) functions converges pointwise to $ u $ in order to see that $ u $ is in fact measurable provided that $ \lbrack u\not=0\rbrack\subseteq\bigcup_{n\in\mathbb N}A_n $). Then $ v:=\overline{\operatorname{sgn} u}\chi_B\in L_{\infty,c}(\mathcal A_c) $. Using part (i) one thus obtains $ c\mu(B)=c\|v\|_1\geq\left|\int_{\Omega}uv\,\mathrm d\mu\right|=\int_{B}|u|\,\mathrm d\mu>c\mu(B) $, which is absurd. 4) In the proof of Theorem 10.15 it is assumed that $ 1<p<\infty $, which seems to be needed in order to apply Exercise 10.3, but afterwards it is not explicitely stated that the subsequent arguments also apply to the case $ p=1 $. In fact, the proof of the strong continuity seems to be most interesting in the case $ p=1 $ as for the remaining cases one could also apply the method used in part (iii) of the proof of Theorem 10.8 based on Hölder's inequality. Moreover, as I understand it, you should consider $ T|_{L_2\cap L_1\cap L_\infty(\mu)} $ in lieu of $ T|_{L_1\cap L_\infty(\mu)} $ within the extension argument. Am I right? 5) Right after the definition of $ L_1(\mathcal A_c) $ it is stated that the elements of $ L_1(\mathcal A_c) $ are to be understood as equivalence classes of a.e. equal functions. In my opinion it would be helpful if you specify in which sense you consider two functions, say $ u $ and $ v $, as a.e. equal: Do you want that $ \lbrack u\not=v\rbrack $ is an element of $ \mathcal A $ with $ \mu $-measure zero or do you want that $ \lbrack u\not=v\rbrack $ is contained in such a set? In the latter case, I think one should be a little bit more careful with statements like $ u\in L_p(\mu) $ as in Lemma 10.7, as in general not all elements of the equivalence class of $ u $ in the latter sense are measurable with respect to $ \mathcal A $ and concerning integration one thus actually works with the completion of $ \mu $. Moreover, I think it would be a little bit more consistent to replace the condition $ \lbrack u\not=0\rbrack\subseteq\bigcup_{n\in\mathbb N}A_n $ in the definition of $ L_1(\mathcal A_c) $ by the requirement that $ \lbrack u\not=0\rbrack\setminus\bigcup_{n\in\mathbb N}A_n $ is (contained in) a $ \mu $-null set or, equivalently, by explicitely demanding that the condition $ \lbrack u\not=0\rbrack\subseteq\bigcup_{n\in\mathbb N}A_n $ has to be satisfied for at least one representative of the equivalence class of $ u $. Best wishes, merry christmas and a happy new year, Heiko
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Dear Heiko, 1) I agree with you concerning the name `three lines theorem' for Cor. 10.3. Theorem 10.1 I would rather call a maximum principle, whereas Phragmen-L. is reserved for Exercise 10.1. (I had had in mind to mention Phragmen-L. somewhere in the lecture, but finally had forgotten.) 2) You are right; we had overlooked that $ {\rm sgn} $ had only been defined for $ \mathbb R $. But the definition in $ \mathbb C $ is the same formula. 3) In view of your counterexample we will certainly not try to give a proof of this statement. I guess we had in mind something like your first proof, but overlooked that it does not work within $ \mathcal A_{\rm c} $. Thanks a lot for pointing out this mistake, and also many thanks for the repair. 4) Our formulation was unfortunate. Starting with the second sentence of (a), everything holds for $ 1\leqslant p<\infty $, and this should have been formulated. You are quite correct that the case $ p=1 $ is the most important, and that then Hölder's inequality could have been used to prove it for the other cases. However, we decided that Lemma 10.16 is quite nice for general p, and as we presented it, the proof is shorter. Concerning your last remark in 4) I mention that $ L_1\cap L_\infty(\mu)\subseteq L_2(\mu) $, so there is no need to add the intersection with $ L_2 $. 5) We had thought about things like you mention. In fact I think everything is correct if interpreted restrictively. The space should be as indicated, and the equivalence classes formed within $ L_1(\mathcal A_{\rm c}) $. These functions are automatically measurable, and therefore $ [u\ne v]\in\mathcal A $, and the problem with non-completeness of the measure space does not come into play. In order to make it clearer, we might have added `(with the equivalence classes formed in $ L_1(\mathcal A_{\rm c}) $)'. Many thanks for your helpful communication; we are very grateful for the remarks. Best wishes, merry christmas and a happy new year, Jürgen
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Dear Heiko, let me add a few things to what Jürgen wrote. 1) I didn't notice this at all -- thanks for pointing it out! (I would also call 10.1 a maximum principle.) 3) Just in case you're wondering why on earth we made this mistake: there was a point in time where this part of the proof was correct. Then we relaxed the conditions on $ \mathcal A_{\rm c} $ and did not see that this part needs to be adapted. What we should have written is that there exists $ A\in\mathcal A $ with $ \mathbf1_A\in L_{\infty,\rm c}(\mathcal A_{\rm c}) $, $ \mu(A)>0 $ and $ A\subseteq[|u|>c] $. (And you gave the full explanation why such a set $ A $ exists, thanks.) 4) The present formulation at the beginning of the proof is just wrong, yes. 5) More strongly than Jürgen wrote I'd say we should have written "(with the equivalence classes formed in $ L_1(\mathcal A_{\rm c}) $)". Best wishes, and a merry christmas, Hendrik
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Dear Jürgen, dear Hendrik, thanks a lot for your additional explanations. Remark 5 of my previous posting was originally motivated by my try to comprehend the remark "We note that for $ \sigma $-finite measure spaces the requirement on $ \lbrack u\not=0\rbrack $ in the definition of $ L_1(\mathcal A_c) $ can be dispensed with." Putting $ \widetilde L_1(\mathcal A_c):=\{u:\Omega\rightarrow\mathbb C;\, \chi_A u\in L_1(\mu)\, \forall A\in\mathcal A_c\} $, I understand the cited remark in the sense that it is claimed that $ \widetilde L_1(\mathcal A_c)=L_1(\mathcal A_c) $ provided that $ (\Omega,\mathcal A,\mu) $ is $ \sigma $-finite. Is that the correct interpretation? If so, then, given the indicated definitions, the claim is wrong in general. For instance, let $ \emptyset\not=N\subseteq\mathbb R\setminus\lbrack 0,1\rbrack $ be a Borel null set and consider $ \Omega=\lbrack 0,1\rbrack\cup N $ with $ \mathcal A $ the Borel $ \sigma $-algebra, $ \mu $ the Lebesgue measure and $ \mathcal A_c $ the ring generated by the left open subintervals of $ \lbrack 0,1\rbrack $. Then $ \chi_N $ clearly belongs to $ \widetilde L_1(\mathcal A_c) $, but not to $ L_1(\mathcal A_c) $. Even worse, choosing $ N $ such that it contains a non-Borel subset $ N' $, the function $ \chi_{N'} $ is a non-measurable element of $ \widetilde L_1(\mathcal A_c) $. In fact, the afore-mentioned claim becomes valid under an additional requirement. More precisely, for a $ \sigma $-finite measure space $ (\Omega,\mathcal A,\mu) $, the subsequent statements are equivalent: (a) $ Omega $ is the countable union of sets belonging to $ \mathcal A_c $; (b) one has $ \widetilde L_1(\mathcal A_c)\subseteq L_1(\mathcal A_c) $; (c) every $ \mu $-null set belonging to $ \mathcal A $ can be covered by the countable union of sets belonging to $ \mathcal A_c $. Indeed, (a) obviously implies (b), while (c) results from (b) due to $ \chi_N\in\widetilde L_1(\mathcal A_c) $ for all $ N\in\mathcal A $ with $ \mu(N)=0 $. Now assume (c). Since $ (\Omega,\mathcal A,\mu) $ is $ \sigma $-finite, it suffices to show that each $ A\in\mathcal A $ of finite $ \mu $-measure can be covered by the countable union of sets belonging to $ \mathcal A_c $. Thanks to the hypotheses, we can find simple functions $ u_n=\sum_{j=1}^{m_n}a_{j,n}\chi_{A_{j,n}} $ with $ n\in\mathbb N $, $ m_n\in\mathbb N $, $ a_{j,n}\in\mathbb C $ and $ A_{j,n}\in\mathcal A_c $ such that $ \lim_{n\to\infty}\|\chi_A-u_n\|_1=0 $. We set $ B:=A\setminus\bigcup_{n\in\mathbb N}\bigcup_{j=1}^{m_n}A_{j,n} $. As we assume (c), it is sufficient to verify that $ B $ is a null set in order to finish the proof. We observe that $ \chi_A\cdot\chi_B=\chi_B $ and $ u_n\cdot\chi_{B}=0 $. We therefore obtain $ \mu(B)=\|\chi_B\|_1=\|\chi_B(\chi_A-u_n)\|_1\leq\|\chi_A-u_n\|_1 $. Now letting $ n $ tend to $ \infty $ gives us $ \mu(B)=0 $. Best wishes, Heiko
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Dear Heiko, once again, thanks for your remarks! You are absolutely correct with you criticism. We simply should have omitted this remark. However, after some more thinking we might come to another opinion and adapt it differently. For an explanation of the history of the proof: We first had a version for the $ \sigma $-finite case, with a countable covering by sets of finite measure and $ \mathcal A_{\rm c} $ the system of measurable subsets of one of these sets. We knew, however, that $ \sigma $-finiteness of the measure space is not really needed, and at a rather late point we (hoped to have) found a way to treat the general case without formulating complicated conditions. Unfortunately, some of the fine points escaped us (or we were too confident and negligent). So, we are all the more grateful for all your critical and contributing comments. Best wishes, Jürgen
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Dear Heiko, after some further thinking we have come to a different opinion. We want to make it correct by two modifications: The first is that in the definition of $ L_1(\mathcal A_{\rm c}) $ we want to add that $ u $ should be measurable. The second is that the equivalence classes should be taken in the measurable functions. Then everything concerning your comment 5) should work, and the comment on the $ \sigma $-finite case will not have to be taken out. Best wishes, Jürgen
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Typos
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Dear all, in Theorem 10.8 the definition of $ \theta_\tau $ should be $ \theta_\tau:=\tau\theta $. Furthermore, in the proof of Theorem 10.9, clearly $ -A $ is m-accretive (instead of $ A $). Best wishes, Christian
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