Timeline for Differentiability along hyperplanes
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12 at 13:18 | comment | added | Jan Bohr | I was trying to see if the example survives strengthening the hypothesis of the question (i.e. differentiable along planes and continuous at zero). The next step would be to ask for a rational counterexample (which might not exist) -- but I'll post this as a separate question. | |
| Jun 12 at 13:13 | comment | added | Saúl RM | Yes, the construction provides a lot of flexibility, you can make it continuous if you want (I preferred to make it discontinuous as that made it, in my opinion, a stronger counterexample with the given hypotheses) | |
| Jun 12 at 13:10 | comment | added | Jan Bohr | Great argument! You can also modify it to be continuous at $0$, by setting $f(x)=\sum_n a_n\phi_n(x)$ for a sequence $a_n\to 0$ and taking $\phi_n(p_n)=1$. Since $N(\epsilon)=\min\{n\in \mathbb N: A_n\cap B(0,\epsilon)\}\to \infty$, as $\epsilon\to 0$, we have $\sup_{|x|<\epsilon}|f(x)|\le a_{N(\epsilon)} \to 0$ as $\epsilon\to 0$, which gives continuity. If $f$ was differentiable at $0$, its differential would have to be zero and thus $b_n=f(p_n)/|p_n|\to 0$. But $b_n=a_n/|p_n|$ and we get a contradiction by choosing e.g. $a_n=|p_n|$. | |
| Jun 12 at 13:05 | vote | accept | Jan Bohr | ||
| Jun 11 at 13:08 | history | answered | Saúl RM | CC BY-SA 4.0 |