Discussion Board Lecture 01

Hi, I believe that if you want to use 1.10(b) for the first equality, you need to interchange the integral with the generator A. For a proof of that I'd use 1.8 (b) (1.8 (a) is not applicable because A is not necessarily continuous). For 1.8 (b) to be true we need that AT(s)x_n is continuous in s which is true, because by 1.10(a) AT(s)x_n=T(s)Ax_n.

Dear Gustavo,

Alexander Kreiß is right, you would need to explain why you can take the $ A $ under the integral, and indeed, that's what Hille's theorem 1.8(b) is good for. However, in order to apply Hille's theorem, you need not only the continuity mentioned by Alexander, but also the closedness of $ A $ (1.10(c)).

The explanation given in parentheses (1.10(a) and 1.9(b)) is the most elementary explanation I can think of. Put in words, it says: use the fundamental theorem of calculus and take into account that we know the derivative of $ t\mapsto T(t)x_n $.

Best wishes,  Hendrik

Dear Gustavo,

for the second equality in the formula one has to use 1.8(a) (applied to the projections), as you say.

Best wishes, Jürgen

Dear all,

I think that it should be sufficient to show that $ \int_0^t T(s) x_n d s \in \mathrm{dom} (\overline{A|_D}) $. But this follows from the first part of Theorem 1.8 (b) as $ \overline{A|_D} $ is closed. As $ \overline{A|_D} $ is a restriction of $ A $, one may use now Theorem 1.10 (b).

Best wishes, André

Dear Hendrik (and others):

Thank you for the clarifications. I made a mistake and missed "taking A under the integral". Correcting myself (with your help), I believe that an alternative justification would be "1.10(b) + 1.10(c) + 1.8(b) for the first equality and 1.8(a) for the second equality". But in fact the justification in the notes is simpler.

Dear André,

I do not understand what you want to say. "Sufficient" for what?

Best wishes, Jürgen

Dear Jürgen,

I meant that $ \int_0^t T(s) x_n d s \in \mathrm{dom} (\overline{A|_D}) $ would be sufficient for $ (\int_0^t T(s) x_n d s , T(t)x_n - x_n ) \in \overline{A|_D} $ as $ \overline{A|_D} $ is a restriction of $ A $ and as we have Theorem 1.10(b). I am sorry if my first comment was not clear.

Best wishes, André

Dear André,

yes, I understand. Nice observation!

Best wishes, Jürgen

Dear Balint,

I created a "general discussions" board. We will try to add a Preview option as soon as possible. One should be able to edit the own comments; this seems to be a bug which we will have to fix (thanks for pointing this out).

Best wishes, Christian

Dear Balint

Posted by Balint Farkas on 21 October 2014 at 08:40.

» even though I was eagerly looking for typos, I have not found any :-) «

Best wishes, Raffael

Dear Balint, dear Raffael,

thanks, Raffael! Actually, there is still another one: In the fourth line of the proof of Theorem 1.10 there is a \to, where there should be a \mapsto.

Best wishes, Jürgen

Dear Andras,

thanks for the super-fast first comment! Of course, the series should be as in Example 1.2. Sorry for the typo! (Unavoidably, there will be others.)

Best wishes, Jürgen

dear Andras, i guess that in the case $ \sum_{n \geq 0} || A^j || $ we need that either the sum is finite (nilpontent for some power $ k $) or the norm of A is strictly less than one, in which case you can bound by the geometric series since $ ||A^j|| \leq ||A||^j $ always holds.

Dear Andras,

I think your (implicit) conjecture was right: If $ e^{A}=\sum_{n\geq 0} A^n $ holds, then $ A^2=0 $. (The converse implication is what you have mentioned explicitly).

Indeed, if the series on the right-hand side converges, then $ A^n\to 0 $ for $ n\to \infty $, so that $ A $ is power-bounded, and we may assume that $ A $ is a strict contraction (renorm the space). From the assumed identity we can conclude $ e^{-A}=I-A $. For $ B=-A $ this yields (by writing up the exponential series) $ 0=\frac{B^2}{2}(I+C) $, where $ C=\frac{2B}{3!}+\frac{2B^2}{4!}+\cdots $. Estimate the norm $ \|C\|\leq 2(e-1-1-\frac{1}2)\leq 0.5<1 $. Hence $ I+C $ is invertible, and we obtain $ B^2=A^2=0 $.

Best wishes,

Balint