Show that $f$ is differentiable on $(0,infty)$ if and only if $f$ is differentiable at $1$












3












$begingroup$


Suppose $f:(0,infty) to mathbb{R}$ satisfies $f(x/y) = f(x) - f(y)$ for all $x,y>0$ and that $f(1) = 0$.



Question: Show that $f$ is differentiable on $(0,infty)$ if and only if $f$ is differentiable at $1$.



My attempt on this problem:



Let $a in (0,infty)$. Then $f'(x) = lim_{hto 0} frac{f(a+h) - f(a)}{h} = lim_{hto 0} frac{f(frac{a+h}{a})}{h} = lim_{hto 0} frac{f(1 +frac{h}{a})}{h}.$



All I know is that the top will go to $f(1) = 0$ and the bottom will also go to $0$ but I don't know how to show that this is differentiable on $(0,infty)$. Any help would be appreciated.






real-analysis limits derivatives







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $f(1)=0$ would help you.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:44






  • 1




    $begingroup$
    $ frac {frac {f(1+frac ha)}a}{frac ha}=frac {f(1+frac ha)}h$ This should also help
    $endgroup$
    – NL1992
    Dec 3 '18 at 4:52












  • $begingroup$
    Thank you guys!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04










  • $begingroup$
    The value of $f(1)$ follows from putting $x=y$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 10:42
















3












$begingroup$


Suppose $f:(0,infty) to mathbb{R}$ satisfies $f(x/y) = f(x) - f(y)$ for all $x,y>0$ and that $f(1) = 0$.



Question: Show that $f$ is differentiable on $(0,infty)$ if and only if $f$ is differentiable at $1$.



My attempt on this problem:



Let $a in (0,infty)$. Then $f'(x) = lim_{hto 0} frac{f(a+h) - f(a)}{h} = lim_{hto 0} frac{f(frac{a+h}{a})}{h} = lim_{hto 0} frac{f(1 +frac{h}{a})}{h}.$



All I know is that the top will go to $f(1) = 0$ and the bottom will also go to $0$ but I don't know how to show that this is differentiable on $(0,infty)$. Any help would be appreciated.






real-analysis limits derivatives







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $f(1)=0$ would help you.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:44






  • 1




    $begingroup$
    $ frac {frac {f(1+frac ha)}a}{frac ha}=frac {f(1+frac ha)}h$ This should also help
    $endgroup$
    – NL1992
    Dec 3 '18 at 4:52












  • $begingroup$
    Thank you guys!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04










  • $begingroup$
    The value of $f(1)$ follows from putting $x=y$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 10:42














3












3








3


1



$begingroup$


Suppose $f:(0,infty) to mathbb{R}$ satisfies $f(x/y) = f(x) - f(y)$ for all $x,y>0$ and that $f(1) = 0$.



Question: Show that $f$ is differentiable on $(0,infty)$ if and only if $f$ is differentiable at $1$.



My attempt on this problem:



Let $a in (0,infty)$. Then $f'(x) = lim_{hto 0} frac{f(a+h) - f(a)}{h} = lim_{hto 0} frac{f(frac{a+h}{a})}{h} = lim_{hto 0} frac{f(1 +frac{h}{a})}{h}.$



All I know is that the top will go to $f(1) = 0$ and the bottom will also go to $0$ but I don't know how to show that this is differentiable on $(0,infty)$. Any help would be appreciated.






real-analysis limits derivatives







share|cite|improve this question









$endgroup$




Suppose $f:(0,infty) to mathbb{R}$ satisfies $f(x/y) = f(x) - f(y)$ for all $x,y>0$ and that $f(1) = 0$.



Question: Show that $f$ is differentiable on $(0,infty)$ if and only if $f$ is differentiable at $1$.



My attempt on this problem:



Let $a in (0,infty)$. Then $f'(x) = lim_{hto 0} frac{f(a+h) - f(a)}{h} = lim_{hto 0} frac{f(frac{a+h}{a})}{h} = lim_{hto 0} frac{f(1 +frac{h}{a})}{h}.$



All I know is that the top will go to $f(1) = 0$ and the bottom will also go to $0$ but I don't know how to show that this is differentiable on $(0,infty)$. Any help would be appreciated.






real-analysis limits derivatives




real-analysis limits derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 4:42









Pika_2018Pika_2018

204




204








  • 2




    $begingroup$
    $f(1)=0$ would help you.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:44






  • 1




    $begingroup$
    $ frac {frac {f(1+frac ha)}a}{frac ha}=frac {f(1+frac ha)}h$ This should also help
    $endgroup$
    – NL1992
    Dec 3 '18 at 4:52












  • $begingroup$
    Thank you guys!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04










  • $begingroup$
    The value of $f(1)$ follows from putting $x=y$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 10:42














  • 2




    $begingroup$
    $f(1)=0$ would help you.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:44






  • 1




    $begingroup$
    $ frac {frac {f(1+frac ha)}a}{frac ha}=frac {f(1+frac ha)}h$ This should also help
    $endgroup$
    – NL1992
    Dec 3 '18 at 4:52












  • $begingroup$
    Thank you guys!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04










  • $begingroup$
    The value of $f(1)$ follows from putting $x=y$.
    $endgroup$
    – Paramanand Singh
    Dec 3 '18 at 10:42








2




2




$begingroup$
$f(1)=0$ would help you.
$endgroup$
– xbh
Dec 3 '18 at 4:44




$begingroup$
$f(1)=0$ would help you.
$endgroup$
– xbh
Dec 3 '18 at 4:44




1




1




$begingroup$
$ frac {frac {f(1+frac ha)}a}{frac ha}=frac {f(1+frac ha)}h$ This should also help
$endgroup$
– NL1992
Dec 3 '18 at 4:52






$begingroup$
$ frac {frac {f(1+frac ha)}a}{frac ha}=frac {f(1+frac ha)}h$ This should also help
$endgroup$
– NL1992
Dec 3 '18 at 4:52














$begingroup$
Thank you guys!
$endgroup$
– Pika_2018
Dec 3 '18 at 5:04




$begingroup$
Thank you guys!
$endgroup$
– Pika_2018
Dec 3 '18 at 5:04












$begingroup$
The value of $f(1)$ follows from putting $x=y$.
$endgroup$
– Paramanand Singh
Dec 3 '18 at 10:42




$begingroup$
The value of $f(1)$ follows from putting $x=y$.
$endgroup$
– Paramanand Singh
Dec 3 '18 at 10:42










1 Answer
1






active

oldest

votes


















4












$begingroup$

If $f$ is differentiable on $(0,infty)$,then $f$ is differentiable at $1$.



For the other direction, suppose $f$ is differentiable at $1$,i.e.$$lim_{hto 0} frac{f(1+h) - f(1)}{h}=lim_{hto 0} frac{f(1+h)-0}{h}$$ exists. Now let $a in (0,infty)$. Then
begin{align}
f'(a) &= lim_{hto 0} frac{f(a+h) - f(a)}{h}\
& = lim_{hto 0} frac{f(frac{a+h}{a})}{h} \
&= lim_{hto 0} frac{f(1 +frac{h}{a})}{h}
end{align}
Let $u=frac{h}{a}$. Then we have
begin{align}
f'(a) &=lim_{uto 0} frac{f(1 +u)}{au}\
&=frac{f'(1)}{a}
end{align}
Since $aneq 0$, $f'(a)$ exists.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you so much!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04











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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

votes









4












$begingroup$

If $f$ is differentiable on $(0,infty)$,then $f$ is differentiable at $1$.



For the other direction, suppose $f$ is differentiable at $1$,i.e.$$lim_{hto 0} frac{f(1+h) - f(1)}{h}=lim_{hto 0} frac{f(1+h)-0}{h}$$ exists. Now let $a in (0,infty)$. Then
begin{align}
f'(a) &= lim_{hto 0} frac{f(a+h) - f(a)}{h}\
& = lim_{hto 0} frac{f(frac{a+h}{a})}{h} \
&= lim_{hto 0} frac{f(1 +frac{h}{a})}{h}
end{align}
Let $u=frac{h}{a}$. Then we have
begin{align}
f'(a) &=lim_{uto 0} frac{f(1 +u)}{au}\
&=frac{f'(1)}{a}
end{align}
Since $aneq 0$, $f'(a)$ exists.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you so much!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04
















4












$begingroup$

If $f$ is differentiable on $(0,infty)$,then $f$ is differentiable at $1$.



For the other direction, suppose $f$ is differentiable at $1$,i.e.$$lim_{hto 0} frac{f(1+h) - f(1)}{h}=lim_{hto 0} frac{f(1+h)-0}{h}$$ exists. Now let $a in (0,infty)$. Then
begin{align}
f'(a) &= lim_{hto 0} frac{f(a+h) - f(a)}{h}\
& = lim_{hto 0} frac{f(frac{a+h}{a})}{h} \
&= lim_{hto 0} frac{f(1 +frac{h}{a})}{h}
end{align}
Let $u=frac{h}{a}$. Then we have
begin{align}
f'(a) &=lim_{uto 0} frac{f(1 +u)}{au}\
&=frac{f'(1)}{a}
end{align}
Since $aneq 0$, $f'(a)$ exists.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you so much!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04














4












4








4





$begingroup$

If $f$ is differentiable on $(0,infty)$,then $f$ is differentiable at $1$.



For the other direction, suppose $f$ is differentiable at $1$,i.e.$$lim_{hto 0} frac{f(1+h) - f(1)}{h}=lim_{hto 0} frac{f(1+h)-0}{h}$$ exists. Now let $a in (0,infty)$. Then
begin{align}
f'(a) &= lim_{hto 0} frac{f(a+h) - f(a)}{h}\
& = lim_{hto 0} frac{f(frac{a+h}{a})}{h} \
&= lim_{hto 0} frac{f(1 +frac{h}{a})}{h}
end{align}
Let $u=frac{h}{a}$. Then we have
begin{align}
f'(a) &=lim_{uto 0} frac{f(1 +u)}{au}\
&=frac{f'(1)}{a}
end{align}
Since $aneq 0$, $f'(a)$ exists.






share|cite|improve this answer









$endgroup$



If $f$ is differentiable on $(0,infty)$,then $f$ is differentiable at $1$.



For the other direction, suppose $f$ is differentiable at $1$,i.e.$$lim_{hto 0} frac{f(1+h) - f(1)}{h}=lim_{hto 0} frac{f(1+h)-0}{h}$$ exists. Now let $a in (0,infty)$. Then
begin{align}
f'(a) &= lim_{hto 0} frac{f(a+h) - f(a)}{h}\
& = lim_{hto 0} frac{f(frac{a+h}{a})}{h} \
&= lim_{hto 0} frac{f(1 +frac{h}{a})}{h}
end{align}
Let $u=frac{h}{a}$. Then we have
begin{align}
f'(a) &=lim_{uto 0} frac{f(1 +u)}{au}\
&=frac{f'(1)}{a}
end{align}
Since $aneq 0$, $f'(a)$ exists.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 '18 at 4:59









Thomas ShelbyThomas Shelby

2,655421




2,655421








  • 1




    $begingroup$
    Thank you so much!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04














  • 1




    $begingroup$
    Thank you so much!
    $endgroup$
    – Pika_2018
    Dec 3 '18 at 5:04








1




1




$begingroup$
Thank you so much!
$endgroup$
– Pika_2018
Dec 3 '18 at 5:04




$begingroup$
Thank you so much!
$endgroup$
– Pika_2018
Dec 3 '18 at 5:04


















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