Timeline for Polar decomposition of the Volterra integral operator
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 24, 2020 at 21:11 | comment | added | ViktorStein | Let us continue this discussion in chat. | |
| Mar 24, 2020 at 21:11 | comment | added | Christian Remling | But of course $\widetilde{K}(x,t) = \int_0^1 K(s,x)K(s,t)\, ds$ must give the same answer, as you pointed out. | |
| Mar 24, 2020 at 21:10 | comment | added | Christian Remling | I think my formula $\widetilde{K}(s,t)=1-\max\{ t,s \}$ is fine. I simply changed the order of integration in $(T^*Tf)(x)=\int_x^1 dt\int_0^t ds\, f(s)$ to obtain this. | |
| Mar 24, 2020 at 20:54 | comment | added | ViktorStein | So shouldn't the kernel of $T^* T$ be $\tilde{K}(x,t) := \int_0^1 K(t,x) K(x,s) ds$, as $$(T^* T f)x = \int_0^1 \tilde{K}(x,t) f(t) dt?$$ | |
| Mar 24, 2020 at 20:36 | comment | added | ViktorStein | @ChristianRemling I know a kernel for $T$ is $K(x,t) := \mathbb{1}_{t \le x}(x,t)$, as in $$(T f)(x) = \int_0^1 K(x,t) f(t) dt.$$ Futhermore, $K(t,x)$ is a kernel for $T^*$: $$(T^* f)(x) = \int_0^1 K(t,x) f(t) dt.$$ This is consistent with the kernel-free expression I found for $T^* T$ above: $$ (T^* T f)(x) = T^*\left(\int_0^1 K(x,s) f(s) ds \right) = \int_0^1 K(t,x) \int_0^1 K(x,s) f(s) ds dt $$ Am I on the right track? Because I think that $\overline{K(t,x)} = K(t,x) \ne K(x,t)$ and $$\int_0^1 K(x,s) K(set) ds = \int_x^t dx = t - x \ne 1 - \max(x,t).$$ | |
| Mar 24, 2020 at 16:30 | comment | added | Christian Remling | I see. In that case, we need a kernel $K(x,t)$ with $\overline{K(t,x)}=K(x,t)$ and $\int_0^1 K(x,s)K(s,t)\, ds = 1-\max \{ x,t\}$ (since this latter expression is the kernel of $T^*T$), and you would then hope to get the right idea by just staring at this long enough (?). | |
| Mar 24, 2020 at 8:45 | comment | added | ViktorStein | @ChristianRemling The reason I hope there exists an explicit answer is that this exercise is taken from this problem set from LMU Munich wherein it states "Important: the full solution here is to produce the explicit "closed" form of the operators $U_T$ and $|T|$. Giving the canonical decomposition of $|V|$ is not enough, show that in fact $|T|$ is an integral operator and $U_T$ is a unitary operator (with an explicit, "easy" form)." | |
| Mar 23, 2020 at 22:12 | history | edited | ViktorStein | CC BY-SA 4.0 |
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| Mar 23, 2020 at 22:03 | history | edited | ViktorStein | CC BY-SA 4.0 |
added 227 characters in body
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| Mar 23, 2020 at 21:56 | history | asked | ViktorStein | CC BY-SA 4.0 |