Discussion Board Lecture 03
Proof of Lumer-Phillips Theorems
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Dear ISem-Team, 1) Due to several references to other Theorems, the proof of Lumer-Phillips Theorem 3.18 appears to be rather short. For the necessity, i understand that Theorem 2.7 gives you the 'm' for 'm-accretive', but why is A accretive in the first place? Would be nice, if someone can write out the 'complete' proof. 2) In the second case of Lumer-Phillips (Theorem 3.22) the aim is to use Theorem 3.14. I don't see why -A is dense defined (dom(A) dense). I would appreciate a quick answer. Thank you, Raphael
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Dear Raphael, concerning 1): Theorem 2.7 gives you also that $ \|(\lambda+A)^{-1}\|\le\frac1\lambda $, and this implies (3.2). Admittedly, the reference to Remark 3.17 is misleading, as there only the other implication is stated. Concerning 2): I think you refer to the sufficiency part. At the beginning of the proof of this part it is stated that $ -A $ generates a $ C_0 $-semigroup, and this implies that $ A $ is densely defined. Thanks for your questions; we will modify our script to make it correct and better understandable. Best wishes, Jürgen
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Thank you for the answer! I was unsure if the other implication was also true and for the second case I simply forgot that a Generator is necessary densely defined. Raphael
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Two things I haven't understood
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Dear all, you may want to read my comment below before you read further. if this forum is still being read by someone I would be grateful for clarifying the following two points: 1) This concerns Remark 3.21 b). I don't understand why the statement in there is true. In particular: If $ \theta $ is some positive angle and $ e^{i\cdot\alpha}A $ is accretive for every angle $ \alpha\in(-\theta,\theta) $ then in particular A is accretive (for $ \alpha=0 $) and thus by Remark 3.21 a) num(A) needs to be contained in the positive real line. But this is more restricive than num(A) is contained in a sector. 2) This is a more general point on contractive semi-groups on Hilbert spaces. Assume $ T $ is a holomorphic semi-group of contractions of angle $ \theta $ with generator -A. Then the restriction of T to the positive real line is a $ C_0 $ semi-group with the same generator -A (this is by the paragraph following the proof of Theorem 3.13). Now by Lumer-Philips (Theorem 3.18) we conclude that A is m-accretive or m-sectorial of angle 0. But this implies by Theorem 3.22 that -A generates a holomorphic semi-group of angle $ \frac{\pi}{2} $. This would proof that any holomorphic semi-group of contractions can be extended to a semi-group on the whole half plane. Is that the case or is there a mistake in my reasoning above? I would be very happy if someone could help me understanding these two points. Thanks and kind regards, Alex
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I am sorry, in fact there is only thing I haven't understood, namely that in the in the definition of accretive there is a real part which isn't there in the numerical range.
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Dear Alex (we are still `listening' on the board, from time to time), first the answer to your last post: If $ A $ is sectorial, then - in particular - its numerical range lies in the right half-plane of $ \mathbb C $, which implies that $ {\rm Re}(Ax\mid x)\ge0 $ for all $ x\in{\rm dom}(A) $. So, sectoriality does have an implication for the real part. In fact, stretching the definition, one could say that - in the case of a complex Hilbert space - accretivity is the same as sectoriality of angle $ \pi/2 $. (But `sectorial' requires an angle less than $ \pi/2 $!) Concerning 1) in your last but one post: Remark 3.21 states that the numerical range of A is contained in the closed right half-plane. (See Remark 3.15(b) for a glimpse what the notation $ [{\rm Re}\geqslant0] $ means.) Concerning 2): I think the main point is the same error (on your side) as in 1) above. Note also that the sector of sectoriality for the operator is complementary to the sector of holomorphy; see Theorem 3.22. Best wishes, Jürgen
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Dear Jürgen, thank you very much for your effort. What I acutally meant with my second post was: "I figured out why I was wrong". I'm sorry for not communicating that well. Thanks again for your reply. Regards, Alex
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m-accretive
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Dear ISem team, perhaps it is worth mentioning, that with this version of m-accretivity an m-accretive operator is necessarily closed. Best wishes, Christoph
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Dear Christoph, we will mention this in the corrected version of the lectures. Thanks for your suggestion. Best wishes, Jürgen
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Comment on proof of Thm 3.7
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Dear ISem team, Thm. 3.7, proof: Since $ f $ is only locally bounded (and you don't know how big the $ r $ is for $ z_0 $ fixed), you need in line 2 of the proof to take either $ r>0 $ small enough or to apply a compactness argument to cover the compact ball $ B[z_0,r] $. Best wishes, Johannes
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Dear Johannes, concerning your first remark, we took it for granted that the reader knows that local boundedness implies boundedness on compact sets (which indeed is proved by a compactness argument). And thanks for communicating the typo. Best wishes, Jürgen
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Remark 3.6 b)
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Dear Isem-Team, it seems to be obvious that $ E $ is not a subspace of the linear operators from $ X $ to $ Y $. The remark might be more interesting if we consider $ E $ as a subset of the dualspace. Probably, the prime for the dualspace is missing. Best regards Jan Heckwolf
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Dear Jan, you are absolutely correct; the prime is missing. In the proof of `(i) implies (ii)' it is stated (correctly) that $ E $ is is a subset of $ \mathcal L(X,Y)' $. The point of the remark should be that $ E $ is not a vector space. Thanks for pointing out the misprint! Best wishes, Jürgen
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Definition of holomorphic semigroups
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Dear Isem-Team, I think that there is a little problem with the definition at pages 28/29 of holomorphic semigroups of the angle $ \theta $ if $ \theta >\frac{\pi}{2} $. If we consider the property (i) for arbitrary $ z_1,z_2 \in \Sigma_{\theta, 0} $, it could happen that the sum $ z_1+z_2 $ is a negative real number and therefore not in the sector where $ T $ is defined. Either, one should require that $ z_1+z_2 $ is also in the sector, such that the left side is well definded, or one should only do the definition for $ \theta \leq \frac{\pi}{2} $, what is justified because we only consider this type later. Best regards, Thomas Eiter
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Dear Thomas, thanks a lot for catching this! In the end it's just a misprint, but it's a bad one. The `opening angle' $ 2\theta $ of the sector can be up to $ \pi $, which means that we should have $ \theta\le\frac\pi2 $. The sector should always be contained in the right half plane. Best wishes, Hendrik
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Dear Thomas, yes, the angle should be $ \leq\pi/2 $. However, if one keeps the definition also for the case $ \theta >\pi/2 $, now assuming that the semigroup property only holds for all $ z_1, z_2\in\Sigma_\theta $ such that $ z_1+z_2\in\Sigma_\theta $, then one can actually prove that the semigroup extends to an entire semigroup (the semigroup property holds again by a an application of the identity theorem). However, if one has an entire semigroup, then the generator is necessarily bounded. And vice versa: every bounded operator generates an entire semigroup (namely, the exponential function $ e^{zA} $. As a conclusion, the following picture may be appropriate: either a holomorphic semigroup is holomorphic on a sector of angle $ \leq\pi/2 $, or it is entire (and the generator is bounded). Best regards, Ralph
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Comment on Theorem 3.2
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Dear ISem-Team Maybe this is just me, but I find it a bit weird to call a proof trivial ((ii) => (iii)) when you are actually using Hahn-Banach there (or am I mistaken?). Best wishes, Raffael
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Dear Raffael, once you know the Hahn-Banach theorem, (ii) => (iii) is trivial :-) On a more serious note, I think I just have to agree with you, in particular since we failed to mention that $ E=X' $ is norming as a last fact before Theorem 3.2. Best wishes, Hendrik
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Dear Isem-Team, certainly, you won't need this in the following, but in order to use the notion of analyticity introduced on page 25, you can extend the list of characterizations in Theorem 3.2 by item (vi) saying that $f$ is analytic as well. I used this characteriszation once and it would be good to have a nice reference for it. All the best, moppi. (PS: Am I mistaken or should one use the prural of Hilbert space at line 3 from below on page 34?)
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Dear Moppi, maybe it's well hidden, but in Remark 3.1(b) it is written that holomorphic functions are analytic. The converse is not mentioned, but easy enough, I think. As for `Hilbert space' near the bottom of page 34, I don't know how to form the prural since I don't know this grammatical form, but I agree that the singular sounds quite strange here :-) Best wishes, Hendrik
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Dear Isem-Team, I'm pondering about a step in the proof of Theorem 3.2, $ (iii) \Rightarrow (iv). $ I know that one can identify $ X $ with its bidual, but the trick that one can embed $ X $ into a subset of its dual is new to me. How exactly does that work? Best regards, Attila
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Dear Attila, just to point this out: one can always identify $ X $ with a subset of its bidual, and by definition $ X $ is reflexive if and only if you can canonically identify $ X $ with the full bidual. As for your question: in the proof of (iii)$ \Rightarrow $(iv), $ X $ is embedded into the dual of a subset $ E $ of its dual! The embedding $ j\colon X\to E' $ is defined by $ j(x) := [E \ni x' \mapsto x'(x)] $. Note that the assumption that $ E $ is almost norming is precisely what is needed for $ j $ to be an embedding. Best wishes, Hendrik
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