Timeline for Absolutely continuity in variation of constant formula
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2016 at 1:55 | vote | accept | Torpedo | ||
| Sep 10, 2016 at 1:55 | vote | accept | Torpedo | ||
| Sep 10, 2016 at 1:55 | |||||
| Sep 10, 2016 at 1:53 | comment | added | Nawaf Bou-Rabee | Ok. I edited my answer based on your feedback. | |
| Sep 10, 2016 at 1:53 | history | edited | Nawaf Bou-Rabee | CC BY-SA 3.0 |
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| Sep 10, 2016 at 1:46 | comment | added | Torpedo | Well for evey $t \ge s \ge 0$ we have $P_{t-s} \in L(H)$ and for almost every $s \in [0,t]$ we have $f(s) \in H$. Hence for almost every $s$ and every $t$ satisfying $t \ge s \ge 0$ we have $||P_{t-s} f(s)|| \le ||P_{t-s}||_{L(H)} || f(s)||$. I am sorry, I seem to be completely blind at the moment. What do you think can go wrong here? | |
| Sep 10, 2016 at 1:40 | comment | added | Nawaf Bou-Rabee | Here: $\| P_{t-s} f(s) \| \le \| P_{t-s} \|_{\mathcal{L}(H)} \| f(s) \|$. (Note that I added the subscript to distinguish the operator from vector norms.) Is this true for all $s \in [0,t]$? If so, then I will edit my answer. | |
| Sep 10, 2016 at 1:37 | comment | added | Torpedo | Where did I use this? | |
| Sep 10, 2016 at 1:34 | comment | added | Nawaf Bou-Rabee | Is $\| f(s) \|$ uniformly bounded by a constant for all $s \in [0,t]$? I don't see that anywhere in the given assumptions. | |
| Sep 9, 2016 at 23:09 | comment | added | Torpedo | I would be very grateful to you if you could help me why you did it that way. | |
| Sep 9, 2016 at 22:44 | comment | added | Torpedo | sorry for asking again, but I looked first at this only on my cell-phone and now I noticed that I do not see why you needed all this: We have $\int_0^t ||P_sA y_0 ||+ ||P_{t-s} f(s)||ds$. What prevents us from estimating this like $||P_sA y_0 || \le ||P_s || \ ||Ay_0|| \le M e^{\omega t} ||Ay_0||$ by the growth condition for semigroups and $||P_{t-s} f(s)|| \le ||P_{t-s}|| \ ||f(s)|| \le M e^{\omega (t-s)} ||f(s)|| \le M e^{\omega t} ||f(s)||.$ Now, it would be sufficient to have $||f( \cdot)|| \in L^1$, right? | |
| Sep 9, 2016 at 20:39 | vote | accept | Torpedo | ||
| Sep 9, 2016 at 23:08 | |||||
| Sep 9, 2016 at 19:15 | history | answered | Nawaf Bou-Rabee | CC BY-SA 3.0 |