Timeline for Does this property implies Lipschitz continuity?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15 at 13:42 | vote | accept | aureliano_buendia | ||
| Mar 15 at 5:38 | comment | added | Fedor Petrov | @user479223 Oops, you are correct, I forgot to check them | |
| S Mar 14 at 23:02 | history | suggested | CommunityBot | CC BY-SA 4.0 |
proper delimiters
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| Mar 14 at 22:33 | review | Suggested edits | |||
| S Mar 14 at 23:02 | |||||
| Mar 14 at 22:14 | comment | added | user479223 | @FedorPetrov Constant functions do not satisfy this property. | |
| Mar 14 at 21:54 | answer | added | KhashF | timeline score: 2 | |
| Mar 14 at 19:12 | comment | added | Fedor Petrov | Much more may be said. For $n=1$ this yields $f(x)=ax+b$ for constant $a, b$, for $n>1$ only constant functions satisfy this property. | |
| Mar 14 at 16:05 | review | Close votes | |||
| Mar 20 at 3:10 | |||||
| Mar 14 at 15:12 | history | asked | aureliano_buendia | CC BY-SA 4.0 |