Timeline for Is the $W^{1, \infty}$ limit of differentiable functions also differentiable?
Current License: CC BY-SA 4.0
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| Jun 9 at 17:28 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jun 4 at 13:06 | comment | added | Pietro Majer | The Cantor function allows nice examples. Since $ c(x)\le x^{\frac{\log3}{\log2}}$ one has $c(x)^2\le x^{\frac{\log4}{\log2}}=o(x)$ as $x\to0$. So the (odd) function $f(x)=c(|x|)^2\text{sgn}(x)$ on $[-1,1]$ has $f’(0)=0$ and $f(\{D_*f=\infty\})=[f(-1), f(1) ]\setminus\{0\}$ . And this way one can easily miss even a countable set of points of $[f(-1), f(1) ]$ | |
| Jun 4 at 12:26 | comment | added | Pietro Majer | The equality can be achieved even if $D_*f<0$ a.e. For instance if $c(x)$ is the Cantor function, one can take $f(x): =2c(x)-x$, so $f’=-1$ a.e., $f([0,1])=[0,1]=f(\{D_*f=+\infty\})$ | |
| Jun 4 at 12:07 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jun 4 at 11:49 | comment | added | Nate River | I wonder if equality is achieved if $Df = 0$ a.e.? Or even iff… if you stipulate that the inequality hold for every interval. | |
| Jun 4 at 11:47 | comment | added | Nate River | Upon rereading the Cantor function example, I realise this result is even cooler than I thought. | |
| Jun 4 at 6:54 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jun 4 at 6:14 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jun 4 at 6:07 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jun 4 at 5:55 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jun 4 at 5:45 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Jun 4 at 5:36 | history | answered | Pietro Majer | CC BY-SA 4.0 |