Timeline for Entry-wise interpolation of operators
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 7, 2018 at 14:29 | comment | added | Mikael de la Salle | Yes, you are correct, I was clumsy in my notation. | |
| Feb 7, 2018 at 12:51 | comment | added | T. Le | Interesting point! I believe in the map $z\mapsto A^{nz}B^{m(1-z)}$, you meant entrywise power and multiplication (which is what I'm interested in). But I think you still need Hadamard product to conclude that $\|A^{\circ n}\|\leq \|A\|^{n}$ (I put the circle in the power to indicate that the power is entry wise). | |
| Feb 7, 2018 at 12:17 | comment | added | Mikael de la Salle | I don't know what other $\varphi$ work, but the case $\varphi(x,y)=x^s y^t$ (for $s,t$ non-negative real numbers) follows from Stein interpolation. If $(s,t)$ can be written as $(n\theta,m(1-\theta))$ for $n,m$ integers ans $\theta \in (0,1)$ (for example if $s,t$ are integers, $\theta= \frac 1 2$ and $n=2s,m=2t$), consider the map $z \mapsto A^{nz} B^{m(1-z)}$. | |
| Feb 7, 2018 at 12:03 | comment | added | T. Le | Yes, the proof you gave is what I have (in addition to a direct computational proof). Thank for the reference on Stein's interpolation method. Do you have any insights on what $\varphi$ works? The case $\varphi(x,y)=x^sy^t$ (where $s,t$ are non-negative integers) doesn't seem to follow from Stein's interpolation, does it? | |
| Feb 7, 2018 at 9:48 | history | answered | Mikael de la Salle | CC BY-SA 3.0 |