Discussion Board Lecture 06
Remark 9.15
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Dear all, In remark 9.15, I think you need, more than the continuos embedding, density of V in H for justifying why the weak convergence in V imply weak convergence in H. In fact, in section 9.4 there where we apply this remark we have this density (a is j-elliptic and j have dense range). Best wishes, Abdallah.
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Sorry, this is for lecture 9. Abdallah Maichine.
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Question on Example 6.16
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First, thank you for these courses. I need the proof of the fact that V is a Hilbert space?
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Hello Abdallahmaichine, isn't the fact that since $ (u|u)_H\le(u,u)_V $ a Cauchy series in $ V $ is also one in $ H $ and the continuity of the scalar product in $ H $ showing the assertion ? (or am I overlooking something ?) Best regards, Christian Schöner
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Hello Christian, yes every Cauchy sequence in V is a Cauchy sequence in H and then converge in H, but us we need convergence by the norm of V. The fact that V is a Hilbert space is equivalent to the closedness of the positive diagonal form a. You can also show that the associate linear diagonal operator is closed. (We lack something more!). Bests regards, Abdallah Maichine.
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Dear Abdallah, you are quite correct: it is not sufficient to have convergence in $ H $; but Christian's hint is the first step in the proof, namely how to get the candidate for the limit of the Cauchy sequence. I will not carry it out here but I suggest you look at the proof of completeness of $ \ell_2 $. This should give you the idea how to do the remaining step. (Anyway, the case you have here is isometrically isomorphic to $ \ell_2 $ with weights; so you could use the completeness of such a space, if you are familiar with this.) Best wishes, Jürgen
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Dear professor Jürgen, Ok, now I understand that the completeness of $\L^{2}(\mu)$, where $\mu$ is the weighted counting measure, and the isometry between H and $\l^{2}$ answers the quedtion. Best regards, Abdallah.
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Misprint on line 3
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Dear all, on line 3 of the lecture one should read $ (I-\Delta_{\rm D})^{-1} $. It is easy to get confused there. $ \Delta_{\rm D} $ is a negative self-adjoint operator, the `good' operator is the negative Dirichlet Laplacian. Best wishes, Jürgen
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Question on Example 6.4
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Dear ISem-Team Firstly, I think there is a typo in the last line of the equivalences (a prime is missing). Secondly, I wonder whether the following argument for $ \bar{A} = -A^* $ works as well: We have $ A \subset -A^* $ as shown in the first part. The same argument (+ a density argument) should work to get $ A^* \subset -A^{**} $, hence $ A \subset -A^* \subset A^{**} = \bar{A} $. Clearly this implies $ \bar{A} = -A^* $. Best wishes, Raffael
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Dear Raffael, you are speaking of the equivalences in Example 6.4. You are correct; the last equivalence should be $ -g'=f $; thanks! Concerning your question: `The same argument' would show that $ A^*|_{C_c^\infty}\subseteq-(A^*|_{C_c^\infty})^* $, and to conclude that this implies that $ A^*\subseteq A^{**} $ you would have to show that $ \overline{A^*|_{C_c^\infty}}=A^* $ (which just amounts to the statement one wants to show). In any case you end up with using the denseness of $ C_c^\infty(\mathbb R) $ in $ H^1(\mathbb R) $. Best wishes, Jürgen
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Dear Jürgen Oh, I see. The thing I called density argument was just what you did below. The argument looked a bit too complicated to me when I first read it, but it was not. Thank you! Best wishes, Raffael
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Question c of exercice 6.3
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The eigenvalues of minus Laplacian in $ [0,\pi ] $ are $ (\lambda_n=n^{2}), n\in\mathbb{N} $ not $ (\lambda_n=n), n\in\mathbb{N} $. There is a mistake on the solution $u$ proposed in the exercise, the pulsation is $ \sqrt{\lambda} $ not $ \lambda $ and then you will one has $ sin(\sqrt{\lambda}\pi)=0 $ then $ \lambda_n=n^{2} $. Abdallah Maichine.
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