product of two no where differentiable is also nowhere differentiable












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 17:26









amWhy

1




1










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


















  • $begingroup$
    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










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