Timeline for Functions that Calculate their $L_p$ Norm
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 9, 2014 at 14:45 | answer | added | Will Sawin | timeline score: 3 | |
| Jul 4, 2014 at 9:55 | comment | added | Sean Eberhard | Not quite the same, but there is a function $f$ defined on some interval such that $f(p) = \|f\|_p^p$ on that interval. Namely, $1/(1-x)$ on $[0,1]$. | |
| Jul 3, 2014 at 22:22 | history | edited | Yemon Choi |
added top level tag and got rid of the mis-used "special functions" tag
|
|
| Jul 3, 2014 at 4:41 | comment | added | Manfred Weis | @ChristianRemling could you please formulate your comment as an answer? Thanks. | |
| Jul 3, 2014 at 4:02 | comment | added | Christian Remling | For example, $f(x)\ge (b-a)^{1/x}\min f$ if $f(p)=\|f\|_p$, which gives a contradiction at an $x$ that realizes the min if $b-a>1$. The other claims follow in the same way. | |
| Jul 3, 2014 at 3:59 | comment | added | Manfred Weis | @ChristianRemling do you know of a proof? | |
| Jul 3, 2014 at 3:46 | comment | added | Christian Remling | In particular, if $b-a>1$, then there are no solutions, and if $b-a=1$, then the solutions are exactly the constants $f=c>0$. | |
| Jul 3, 2014 at 3:42 | comment | added | Christian Remling | It's easy to see that non-constant examples are only possible if $b-a<1$ and $f\notin L^b$. (I had this posted as what I thought was a complete answer when I discovered that I hadn't read your question carefully enough.) | |
| Jul 2, 2014 at 23:48 | answer | added | Christian Remling | timeline score: 3 | |
| Jul 2, 2014 at 18:56 | history | edited | Pietro Majer | CC BY-SA 3.0 |
added 8 characters in body
|
| Jul 2, 2014 at 13:30 | history | edited | Manfred Weis | CC BY-SA 3.0 |
fixed bracketing
|
| Jul 2, 2014 at 13:18 | history | edited | Manfred Weis | CC BY-SA 3.0 |
added the request for non-constant functions
|
| Jul 2, 2014 at 13:12 | comment | added | Manfred Weis | @MarkMeckes your example is correct; the question would be whether also non-constant functions exist, that calculate their $L_p$ norm | |
| Jul 2, 2014 at 12:55 | comment | added | Mark Meckes | How about $a=0$, $b=1$, $f(x) = 1$? | |
| Jul 2, 2014 at 12:36 | history | asked | Manfred Weis | CC BY-SA 3.0 |