Discussion Board Lecture 05
Question on the theorem 5.9
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Dear ISEM team, I've remarked that in the final part of the statement of the theorem 5.9 is written the operator A is m-sectorial, i.e., -A generates a holomorphic C_0-semigroup on H which is contractive on a sector. It seems that the semigroup is holomorphic on all C, and contractive on a sector; but I think that, by definition, a holomorphic semigroup is always holomorphic on a sector; I think that it's better to write The operator A is m-sectorial, i.e., -A generates a contractive holomorphic C_0-semigroup. Best regards Giorgio Menegatti
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Dear Giorgio, you are right; I agree with you, that it would be better to write it as you say. (Nevertheless, how we have written it, it is not incorrect. Our formulation does not imply that the semigroup is holomorphic on $ \mathbb C $.) Best wishes, Jürgen
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Remarks from Darmstadt
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Dear IsemTeam, We have two little comments regarding Lecture 5. In example 5.10, the denseness of the embedding j is stated but not presented. We discussed the following argument. Let $ \varepsilon > 0, u \in L_2(\Omega, \mu) $ and $ u_{\varepsilon} := \frac{u}{1 + \varepsilon \sqrt{w}} $. Then $ u_{\varepsilon} \in L_2(\Omega,w \mu) $ and $ u_{\varepsilon} \underset{\varepsilon \to 0}\to u $. The first statement is true because $ w \left|u_{\varepsilon}^2 \right| = \frac{w |u|^2}{(1 + \varepsilon \sqrt{w})^2} \leq \frac{|u^2|}{\varepsilon^2} $ and hence, $ u_{\varepsilon} \in L_2(\Omega,w \mu) $. The second statement is true with the dominated convergence theorem and the upper bound $ |u| $ as $ \left| \frac{u}{1 + \varepsilon \sqrt{w}} \right| \leq |u| $. Is there an easier argument to observe the denseness of the embedding j? In exercise 5.3, the norm to use for the space $ H^2(a,b) $ is not given and it has not been introduced so far. We assume that the usual norm, i.e., the square root of the sum of the squared L2-norms of the partial derivatives up to order two, is the appropriate choice. Cheers, The Darmstadt Team
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Dear Darmstadt Team, dear Paul, yes, there is an easier argument, namely: use Exercise 1.3(b) and note that $ V $ is the domain of the maximal multiplication operator by the function $ \sqrt w $. It would have been nice had we referred to Exercise 1.3 ... As for your second comment, you're right, we forgot to state what the norm is, and it should indeed be the one you said. Thanks a lot for pointing this out! Best wishes, Hendrik
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Dear all, one may also proceed similarly as in Example 5.14: Pick $ f\in j(V)^\perp $. For each $ u\in V $ we then obtain $ 0=(f|j(u))_H=(f/w|u)_V $ (note that $ f/w $ belongs to $ V $ due to $ |f/w|^2w=|f|^2/w\leq|f|^2/\delta $). Hence, $ f/w=0 $ a.e. with respect to $ w\mu $. Because of $ \int_A w\,\mathrm d\mu\geq\delta\mu(A) $ for each set $ A $ in the domain of $ \mu $, we infer $ f=0 $ $ \mu $-almost everywhere. Best wishes, Heiko
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2 small typos
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Dear ISem Team, thanks for the interesting lecture! Let me mention two typos: Example 5.12, line 4: elliptic Last but one sentence in the Notes: It appears that the middle of the sentence is deleted by accident. Best wishes Balint
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Dear Balint, thanks for the remarks! Here is the complete last but one sentence: The reason is that a form $ a\colon V\times V\to\C $ can easily be restricted to a finite-dimensional subspace $ V_m\times V_m $, whereas for operators there might be only few invariant subspaces. Best wishes, Jürgen
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