product of two no where differentiable is also nowhere differentiable
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Is product of two no where differentiable functions always no where differentiable?
For example, we have $f(z) =overline{z} $
and $g(z) = overline{z}^2$.
We know that $f(z)$ is no where differentiable in $Bbb C,$
but from here we find that $g(z)$ is no where differentiable except at zero
so in general, can we say that the product of two non differentiable functions may be differentuable at some point in $Bbb C$ ?
I mean I'm asking, can we conclude any general result ?
complex-analysis
edited Jan 7 at 17:26
amWhy
1
asked Jan 7 at 17:03
HenryHenry
327
$endgroup$
add a comment |
$begingroup$
Is product of two no where differentiable functions always no where differentiable?
For example, we have $f(z) =overline{z} $
and $g(z) = overline{z}^2$.
We know that $f(z)$ is no where differentiable in $Bbb C,$
but from here we find that $g(z)$ is no where differentiable except at zero
so in general, can we say that the product of two non differentiable functions may be differentuable at some point in $Bbb C$ ?
I mean I'm asking, can we conclude any general result ?
complex-analysis
edited Jan 7 at 17:26
amWhy
1
asked Jan 7 at 17:03
HenryHenry
327
$endgroup$
$begingroup$
Duplicate of math.stackexchange.com/questions/948421/… ?
$endgroup$
– Martin R
Jan 7 at 17:13
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Do the functions have to be continuous or have some other properties? If not, why not just partition the domain into two dense sets, $E$ and $F,$ and let $f(x)=1$ for all $x in E$ and $f(x) = -1$ for all $x in F,$ and let $g(x)=-1$ for all $x in E$ and $g(x) = 1$ for all $x in F?$ Then each of $f$ and $g$ is nowhere differentiable, and their product is a constant function.
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– Dave L. Renfro
Jan 7 at 17:44
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Nope. $chi_{[0,1]capmathbb{Q}}cdotchi_{[0,1]capoverline{mathbb{Q}}}=0.$
$endgroup$
– Adrian Keister
Jan 7 at 18:22
add a comment |
$begingroup$
Is product of two no where differentiable functions always no where differentiable?
For example, we have $f(z) =overline{z} $
and $g(z) = overline{z}^2$.
We know that $f(z)$ is no where differentiable in $Bbb C,$
but from here we find that $g(z)$ is no where differentiable except at zero
so in general, can we say that the product of two non differentiable functions may be differentuable at some point in $Bbb C$ ?
I mean I'm asking, can we conclude any general result ?
complex-analysis
edited Jan 7 at 17:26
amWhy
1
asked Jan 7 at 17:03
HenryHenry
327
$endgroup$
Is product of two no where differentiable functions always no where differentiable?
For example, we have $f(z) =overline{z} $
and $g(z) = overline{z}^2$.
We know that $f(z)$ is no where differentiable in $Bbb C,$
but from here we find that $g(z)$ is no where differentiable except at zero
so in general, can we say that the product of two non differentiable functions may be differentuable at some point in $Bbb C$ ?
I mean I'm asking, can we conclude any general result ?
complex-analysis
complex-analysis
edited Jan 7 at 17:26
amWhy
1
asked Jan 7 at 17:03
HenryHenry
327
edited Jan 7 at 17:26
amWhy
1
asked Jan 7 at 17:03
HenryHenry
327
edited Jan 7 at 17:26
amWhy
1
edited Jan 7 at 17:26
amWhy
1
edited Jan 7 at 17:26
amWhy
1
1
asked Jan 7 at 17:03
HenryHenry
327
asked Jan 7 at 17:03
HenryHenry
327
asked Jan 7 at 17:03
HenryHenry
327
327
$begingroup$
Duplicate of math.stackexchange.com/questions/948421/… ?
$endgroup$
– Martin R
Jan 7 at 17:13
$begingroup$
Do the functions have to be continuous or have some other properties? If not, why not just partition the domain into two dense sets, $E$ and $F,$ and let $f(x)=1$ for all $x in E$ and $f(x) = -1$ for all $x in F,$ and let $g(x)=-1$ for all $x in E$ and $g(x) = 1$ for all $x in F?$ Then each of $f$ and $g$ is nowhere differentiable, and their product is a constant function.
$endgroup$
– Dave L. Renfro
Jan 7 at 17:44
$begingroup$
Nope. $chi_{[0,1]capmathbb{Q}}cdotchi_{[0,1]capoverline{mathbb{Q}}}=0.$
$endgroup$
– Adrian Keister
Jan 7 at 18:22
add a comment |
$begingroup$
Duplicate of math.stackexchange.com/questions/948421/… ?
$endgroup$
– Martin R
Jan 7 at 17:13
$begingroup$
Do the functions have to be continuous or have some other properties? If not, why not just partition the domain into two dense sets, $E$ and $F,$ and let $f(x)=1$ for all $x in E$ and $f(x) = -1$ for all $x in F,$ and let $g(x)=-1$ for all $x in E$ and $g(x) = 1$ for all $x in F?$ Then each of $f$ and $g$ is nowhere differentiable, and their product is a constant function.
$endgroup$
– Dave L. Renfro
Jan 7 at 17:44
$begingroup$
Nope. $chi_{[0,1]capmathbb{Q}}cdotchi_{[0,1]capoverline{mathbb{Q}}}=0.$
$endgroup$
– Adrian Keister
Jan 7 at 18:22
$begingroup$
Duplicate of math.stackexchange.com/questions/948421/… ?
$endgroup$
– Martin R
Jan 7 at 17:13
$begingroup$
Duplicate of math.stackexchange.com/questions/948421/… ?
$endgroup$
– Martin R
Jan 7 at 17:13
$begingroup$
Do the functions have to be continuous or have some other properties? If not, why not just partition the domain into two dense sets, $E$ and $F,$ and let $f(x)=1$ for all $x in E$ and $f(x) = -1$ for all $x in F,$ and let $g(x)=-1$ for all $x in E$ and $g(x) = 1$ for all $x in F?$ Then each of $f$ and $g$ is nowhere differentiable, and their product is a constant function.
$endgroup$
– Dave L. Renfro
Jan 7 at 17:44
$begingroup$
Do the functions have to be continuous or have some other properties? If not, why not just partition the domain into two dense sets, $E$ and $F,$ and let $f(x)=1$ for all $x in E$ and $f(x) = -1$ for all $x in F,$ and let $g(x)=-1$ for all $x in E$ and $g(x) = 1$ for all $x in F?$ Then each of $f$ and $g$ is nowhere differentiable, and their product is a constant function.
$endgroup$
– Dave L. Renfro
Jan 7 at 17:44
$begingroup$
Nope. $chi_{[0,1]capmathbb{Q}}cdotchi_{[0,1]capoverline{mathbb{Q}}}=0.$
$endgroup$
– Adrian Keister
Jan 7 at 18:22
$begingroup$
Nope. $chi_{[0,1]capmathbb{Q}}cdotchi_{[0,1]capoverline{mathbb{Q}}}=0.$
$endgroup$
– Adrian Keister
Jan 7 at 18:22
add a comment |
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$begingroup$
Duplicate of math.stackexchange.com/questions/948421/… ?
$endgroup$
– Martin R
Jan 7 at 17:13
$begingroup$
Do the functions have to be continuous or have some other properties? If not, why not just partition the domain into two dense sets, $E$ and $F,$ and let $f(x)=1$ for all $x in E$ and $f(x) = -1$ for all $x in F,$ and let $g(x)=-1$ for all $x in E$ and $g(x) = 1$ for all $x in F?$ Then each of $f$ and $g$ is nowhere differentiable, and their product is a constant function.
$endgroup$
– Dave L. Renfro
Jan 7 at 17:44
$begingroup$
Nope. $chi_{[0,1]capmathbb{Q}}cdotchi_{[0,1]capoverline{mathbb{Q}}}=0.$
$endgroup$
– Adrian Keister
Jan 7 at 18:22