If $f$ is a derivative then is $|f|$ also a derivative?












2












$begingroup$


If $f$ is a derivative then, is $|f|$ also a derivative?



If $f$ is Riemann integrable then it's true. But, if it's not the case,then is it true?






riemann-integration







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$endgroup$












  • $begingroup$
    By saying, $f$ is a derivative, I want to mean that there exists a function $g$ such that $g'(x)=f(x)$
    $endgroup$
    – Tom.
    Jan 24 at 4:40










  • $begingroup$
    So, technically, a function is a derivative iff $int_0^x f(t)dt= g(x)$ exists for all $x$ and is differentiable everywhere. Is that it?
    $endgroup$
    – stressed out
    Jan 24 at 4:41


















2












$begingroup$


If $f$ is a derivative then, is $|f|$ also a derivative?



If $f$ is Riemann integrable then it's true. But, if it's not the case,then is it true?






riemann-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    By saying, $f$ is a derivative, I want to mean that there exists a function $g$ such that $g'(x)=f(x)$
    $endgroup$
    – Tom.
    Jan 24 at 4:40










  • $begingroup$
    So, technically, a function is a derivative iff $int_0^x f(t)dt= g(x)$ exists for all $x$ and is differentiable everywhere. Is that it?
    $endgroup$
    – stressed out
    Jan 24 at 4:41
















2












2








2





$begingroup$


If $f$ is a derivative then, is $|f|$ also a derivative?



If $f$ is Riemann integrable then it's true. But, if it's not the case,then is it true?






riemann-integration







share|cite|improve this question









$endgroup$




If $f$ is a derivative then, is $|f|$ also a derivative?



If $f$ is Riemann integrable then it's true. But, if it's not the case,then is it true?






riemann-integration




riemann-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 4:23









Tom.Tom.

15619




15619












  • $begingroup$
    By saying, $f$ is a derivative, I want to mean that there exists a function $g$ such that $g'(x)=f(x)$
    $endgroup$
    – Tom.
    Jan 24 at 4:40










  • $begingroup$
    So, technically, a function is a derivative iff $int_0^x f(t)dt= g(x)$ exists for all $x$ and is differentiable everywhere. Is that it?
    $endgroup$
    – stressed out
    Jan 24 at 4:41




















  • $begingroup$
    By saying, $f$ is a derivative, I want to mean that there exists a function $g$ such that $g'(x)=f(x)$
    $endgroup$
    – Tom.
    Jan 24 at 4:40










  • $begingroup$
    So, technically, a function is a derivative iff $int_0^x f(t)dt= g(x)$ exists for all $x$ and is differentiable everywhere. Is that it?
    $endgroup$
    – stressed out
    Jan 24 at 4:41


















$begingroup$
By saying, $f$ is a derivative, I want to mean that there exists a function $g$ such that $g'(x)=f(x)$
$endgroup$
– Tom.
Jan 24 at 4:40




$begingroup$
By saying, $f$ is a derivative, I want to mean that there exists a function $g$ such that $g'(x)=f(x)$
$endgroup$
– Tom.
Jan 24 at 4:40












$begingroup$
So, technically, a function is a derivative iff $int_0^x f(t)dt= g(x)$ exists for all $x$ and is differentiable everywhere. Is that it?
$endgroup$
– stressed out
Jan 24 at 4:41






$begingroup$
So, technically, a function is a derivative iff $int_0^x f(t)dt= g(x)$ exists for all $x$ and is differentiable everywhere. Is that it?
$endgroup$
– stressed out
Jan 24 at 4:41












2 Answers
2






active

oldest

votes


















4












$begingroup$

This is not true even if $f$ is integrable.



Consider $f(x) = begin{cases}cos(x^{-1}), & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$ and $g(x) = |f|(x) = begin{cases}|cos(x^{-1})|, & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$



We can show that $f = F'$ where $F(x) = int_0^x f(t) , dt$. This is easy to show for $0 < x leqslant 1$ where $f$ is continuous. It is also true but harder to show that $F'(0) = f(0) = 0$.



Now if $g = G'$ on $[0,1]$ and $g$ is integrable then $int_0^x g(t) , dt = G(x)$ by the FTC with $G'(0) = g(0)$.



However,



$$G'(0) = lim_{h to 0} frac{1}{h}int_0^h g(t) , dt = lim_{h to 0} frac{1}{h}int_0^h |cos(t^{-1})| , dt = frac{2}{pi} neq g(0)$$



See here for a proof that $G'(0) = 2/pi$.



Addendum



To prove that $F'(0) = 0$, note that $F(0) = 0$ and with a variable change $y = t^{-1}$,



$$F(x) = int_0^x cos (t^{-1}) , dt = int_{x^{-1}}^infty frac{cos y}{y^2} , dy$$



Integrating by parts we get,



$$F(x) = frac{sin(x^{-1})}{x^{-2}} + 2 int_{x^{-1}}^infty frac{sin y}{y^3} , dy$$



Thus,



$$0 leqslant left|frac{F(x)-F(0)}{x} right| leqslant frac{|sin(x^{-1})|}{x^{-1}} + frac{2}{x} int_{x^{-1}}^infty frac{|sin y|}{y^3} , dy \ leqslant x + frac{2}{x} int_{x^{-1}}^infty frac{1}{y^3} , dy = 2x$$



By the squeeze theorem it follows that



$$F'(0) = lim_{x searrow 0}frac{F(x)-F(0)}{x} = 0 $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Well, I'm replying to your comment here. Yes, I agree with you. All I was saying is that the OP hadn't worded the question as precisely as possible. Anyway, +1 for your answer.
    $endgroup$
    – stressed out
    Jan 24 at 6:31






  • 1




    $begingroup$
    @stressedout: Thank you. It would be appropriate for OP to respond to your request for clarification. Although think the first comment does it for me.
    $endgroup$
    – RRL
    Jan 24 at 6:33












  • $begingroup$
    @RRL, can you give a proof of your claim i.e. $F'(0) = f(0) = 0$?
    $endgroup$
    – Tom.
    Jan 24 at 8:19










  • $begingroup$
    @Tom: I added the proof of that. Note that functions which are derivatives on an interval $[a,b]$ are said to belong to the space $mathcal{D}^1([a,b])$ and I am sure (aside from the counterexample I came up with) that $f in mathcal{D}^1([a,b])$ does not imply $|f| in mathcal{D}^1([a,b])$, but I have not found a reference yet.
    $endgroup$
    – RRL
    Jan 24 at 8:51








  • 1




    $begingroup$
    @Tom: If $f$ is Riemann integrable then $g = |f|$ is always Riemann integrable. In my example $f$ is Riemann integrable. I don’t know what you mean by “I assumed”. In general if $g$ is a derivative it does not have to be integrable. Look up Volterra’s function.
    $endgroup$
    – RRL
    Jan 24 at 15:09



















0












$begingroup$

Consider $f(x)=x$ which is differentiable everywhere but $|f(x)|$ is not differentiable at zero.






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$endgroup$









  • 1




    $begingroup$
    Did you answer the question?
    $endgroup$
    – Jacky Chong
    Jan 24 at 4:48










  • $begingroup$
    It's not your fault. The question is ambiguous. You haven't actually answered the question. Check the comments.
    $endgroup$
    – stressed out
    Jan 24 at 4:50






  • 1




    $begingroup$
    @stressedout: The question is not ambiguous. This answer is about differentiability of $|f|$ -- which does not answer the OP. OP is asking if $f = F'$ on some domain does that mean that $|f| = G'$ for some differentiable function $G$.
    $endgroup$
    – RRL
    Jan 24 at 5:50










  • $begingroup$
    @RRL The question is ambiguous in the sense that the OP has not defined what a 'derivative' function is. I left a comment under their post asking for more explanation because it sounded ambiguous to me at first. Plus, if $f$ is not Riemann integrable, then I don't see how $f=F'$ can hold in the first place. Because if it does, then $int f = f$ exists. No?
    $endgroup$
    – stressed out
    Jan 24 at 5:55












  • $begingroup$
    @stressedout: Good you asked for clarification.
    $endgroup$
    – RRL
    Jan 24 at 6:02











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2 Answers
2






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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

This is not true even if $f$ is integrable.



Consider $f(x) = begin{cases}cos(x^{-1}), & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$ and $g(x) = |f|(x) = begin{cases}|cos(x^{-1})|, & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$



We can show that $f = F'$ where $F(x) = int_0^x f(t) , dt$. This is easy to show for $0 < x leqslant 1$ where $f$ is continuous. It is also true but harder to show that $F'(0) = f(0) = 0$.



Now if $g = G'$ on $[0,1]$ and $g$ is integrable then $int_0^x g(t) , dt = G(x)$ by the FTC with $G'(0) = g(0)$.



However,



$$G'(0) = lim_{h to 0} frac{1}{h}int_0^h g(t) , dt = lim_{h to 0} frac{1}{h}int_0^h |cos(t^{-1})| , dt = frac{2}{pi} neq g(0)$$



See here for a proof that $G'(0) = 2/pi$.



Addendum



To prove that $F'(0) = 0$, note that $F(0) = 0$ and with a variable change $y = t^{-1}$,



$$F(x) = int_0^x cos (t^{-1}) , dt = int_{x^{-1}}^infty frac{cos y}{y^2} , dy$$



Integrating by parts we get,



$$F(x) = frac{sin(x^{-1})}{x^{-2}} + 2 int_{x^{-1}}^infty frac{sin y}{y^3} , dy$$



Thus,



$$0 leqslant left|frac{F(x)-F(0)}{x} right| leqslant frac{|sin(x^{-1})|}{x^{-1}} + frac{2}{x} int_{x^{-1}}^infty frac{|sin y|}{y^3} , dy \ leqslant x + frac{2}{x} int_{x^{-1}}^infty frac{1}{y^3} , dy = 2x$$



By the squeeze theorem it follows that



$$F'(0) = lim_{x searrow 0}frac{F(x)-F(0)}{x} = 0 $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Well, I'm replying to your comment here. Yes, I agree with you. All I was saying is that the OP hadn't worded the question as precisely as possible. Anyway, +1 for your answer.
    $endgroup$
    – stressed out
    Jan 24 at 6:31






  • 1




    $begingroup$
    @stressedout: Thank you. It would be appropriate for OP to respond to your request for clarification. Although think the first comment does it for me.
    $endgroup$
    – RRL
    Jan 24 at 6:33












  • $begingroup$
    @RRL, can you give a proof of your claim i.e. $F'(0) = f(0) = 0$?
    $endgroup$
    – Tom.
    Jan 24 at 8:19










  • $begingroup$
    @Tom: I added the proof of that. Note that functions which are derivatives on an interval $[a,b]$ are said to belong to the space $mathcal{D}^1([a,b])$ and I am sure (aside from the counterexample I came up with) that $f in mathcal{D}^1([a,b])$ does not imply $|f| in mathcal{D}^1([a,b])$, but I have not found a reference yet.
    $endgroup$
    – RRL
    Jan 24 at 8:51








  • 1




    $begingroup$
    @Tom: If $f$ is Riemann integrable then $g = |f|$ is always Riemann integrable. In my example $f$ is Riemann integrable. I don’t know what you mean by “I assumed”. In general if $g$ is a derivative it does not have to be integrable. Look up Volterra’s function.
    $endgroup$
    – RRL
    Jan 24 at 15:09
















4












$begingroup$

This is not true even if $f$ is integrable.



Consider $f(x) = begin{cases}cos(x^{-1}), & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$ and $g(x) = |f|(x) = begin{cases}|cos(x^{-1})|, & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$



We can show that $f = F'$ where $F(x) = int_0^x f(t) , dt$. This is easy to show for $0 < x leqslant 1$ where $f$ is continuous. It is also true but harder to show that $F'(0) = f(0) = 0$.



Now if $g = G'$ on $[0,1]$ and $g$ is integrable then $int_0^x g(t) , dt = G(x)$ by the FTC with $G'(0) = g(0)$.



However,



$$G'(0) = lim_{h to 0} frac{1}{h}int_0^h g(t) , dt = lim_{h to 0} frac{1}{h}int_0^h |cos(t^{-1})| , dt = frac{2}{pi} neq g(0)$$



See here for a proof that $G'(0) = 2/pi$.



Addendum



To prove that $F'(0) = 0$, note that $F(0) = 0$ and with a variable change $y = t^{-1}$,



$$F(x) = int_0^x cos (t^{-1}) , dt = int_{x^{-1}}^infty frac{cos y}{y^2} , dy$$



Integrating by parts we get,



$$F(x) = frac{sin(x^{-1})}{x^{-2}} + 2 int_{x^{-1}}^infty frac{sin y}{y^3} , dy$$



Thus,



$$0 leqslant left|frac{F(x)-F(0)}{x} right| leqslant frac{|sin(x^{-1})|}{x^{-1}} + frac{2}{x} int_{x^{-1}}^infty frac{|sin y|}{y^3} , dy \ leqslant x + frac{2}{x} int_{x^{-1}}^infty frac{1}{y^3} , dy = 2x$$



By the squeeze theorem it follows that



$$F'(0) = lim_{x searrow 0}frac{F(x)-F(0)}{x} = 0 $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Well, I'm replying to your comment here. Yes, I agree with you. All I was saying is that the OP hadn't worded the question as precisely as possible. Anyway, +1 for your answer.
    $endgroup$
    – stressed out
    Jan 24 at 6:31






  • 1




    $begingroup$
    @stressedout: Thank you. It would be appropriate for OP to respond to your request for clarification. Although think the first comment does it for me.
    $endgroup$
    – RRL
    Jan 24 at 6:33












  • $begingroup$
    @RRL, can you give a proof of your claim i.e. $F'(0) = f(0) = 0$?
    $endgroup$
    – Tom.
    Jan 24 at 8:19










  • $begingroup$
    @Tom: I added the proof of that. Note that functions which are derivatives on an interval $[a,b]$ are said to belong to the space $mathcal{D}^1([a,b])$ and I am sure (aside from the counterexample I came up with) that $f in mathcal{D}^1([a,b])$ does not imply $|f| in mathcal{D}^1([a,b])$, but I have not found a reference yet.
    $endgroup$
    – RRL
    Jan 24 at 8:51








  • 1




    $begingroup$
    @Tom: If $f$ is Riemann integrable then $g = |f|$ is always Riemann integrable. In my example $f$ is Riemann integrable. I don’t know what you mean by “I assumed”. In general if $g$ is a derivative it does not have to be integrable. Look up Volterra’s function.
    $endgroup$
    – RRL
    Jan 24 at 15:09














4












4








4





$begingroup$

This is not true even if $f$ is integrable.



Consider $f(x) = begin{cases}cos(x^{-1}), & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$ and $g(x) = |f|(x) = begin{cases}|cos(x^{-1})|, & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$



We can show that $f = F'$ where $F(x) = int_0^x f(t) , dt$. This is easy to show for $0 < x leqslant 1$ where $f$ is continuous. It is also true but harder to show that $F'(0) = f(0) = 0$.



Now if $g = G'$ on $[0,1]$ and $g$ is integrable then $int_0^x g(t) , dt = G(x)$ by the FTC with $G'(0) = g(0)$.



However,



$$G'(0) = lim_{h to 0} frac{1}{h}int_0^h g(t) , dt = lim_{h to 0} frac{1}{h}int_0^h |cos(t^{-1})| , dt = frac{2}{pi} neq g(0)$$



See here for a proof that $G'(0) = 2/pi$.



Addendum



To prove that $F'(0) = 0$, note that $F(0) = 0$ and with a variable change $y = t^{-1}$,



$$F(x) = int_0^x cos (t^{-1}) , dt = int_{x^{-1}}^infty frac{cos y}{y^2} , dy$$



Integrating by parts we get,



$$F(x) = frac{sin(x^{-1})}{x^{-2}} + 2 int_{x^{-1}}^infty frac{sin y}{y^3} , dy$$



Thus,



$$0 leqslant left|frac{F(x)-F(0)}{x} right| leqslant frac{|sin(x^{-1})|}{x^{-1}} + frac{2}{x} int_{x^{-1}}^infty frac{|sin y|}{y^3} , dy \ leqslant x + frac{2}{x} int_{x^{-1}}^infty frac{1}{y^3} , dy = 2x$$



By the squeeze theorem it follows that



$$F'(0) = lim_{x searrow 0}frac{F(x)-F(0)}{x} = 0 $$






share|cite|improve this answer











$endgroup$



This is not true even if $f$ is integrable.



Consider $f(x) = begin{cases}cos(x^{-1}), & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$ and $g(x) = |f|(x) = begin{cases}|cos(x^{-1})|, & 0 < x leqslant 1 \ 0, & x= 0 end{cases}$



We can show that $f = F'$ where $F(x) = int_0^x f(t) , dt$. This is easy to show for $0 < x leqslant 1$ where $f$ is continuous. It is also true but harder to show that $F'(0) = f(0) = 0$.



Now if $g = G'$ on $[0,1]$ and $g$ is integrable then $int_0^x g(t) , dt = G(x)$ by the FTC with $G'(0) = g(0)$.



However,



$$G'(0) = lim_{h to 0} frac{1}{h}int_0^h g(t) , dt = lim_{h to 0} frac{1}{h}int_0^h |cos(t^{-1})| , dt = frac{2}{pi} neq g(0)$$



See here for a proof that $G'(0) = 2/pi$.



Addendum



To prove that $F'(0) = 0$, note that $F(0) = 0$ and with a variable change $y = t^{-1}$,



$$F(x) = int_0^x cos (t^{-1}) , dt = int_{x^{-1}}^infty frac{cos y}{y^2} , dy$$



Integrating by parts we get,



$$F(x) = frac{sin(x^{-1})}{x^{-2}} + 2 int_{x^{-1}}^infty frac{sin y}{y^3} , dy$$



Thus,



$$0 leqslant left|frac{F(x)-F(0)}{x} right| leqslant frac{|sin(x^{-1})|}{x^{-1}} + frac{2}{x} int_{x^{-1}}^infty frac{|sin y|}{y^3} , dy \ leqslant x + frac{2}{x} int_{x^{-1}}^infty frac{1}{y^3} , dy = 2x$$



By the squeeze theorem it follows that



$$F'(0) = lim_{x searrow 0}frac{F(x)-F(0)}{x} = 0 $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 24 at 8:57

























answered Jan 24 at 5:48









RRLRRL

52.3k42573




52.3k42573












  • $begingroup$
    Well, I'm replying to your comment here. Yes, I agree with you. All I was saying is that the OP hadn't worded the question as precisely as possible. Anyway, +1 for your answer.
    $endgroup$
    – stressed out
    Jan 24 at 6:31






  • 1




    $begingroup$
    @stressedout: Thank you. It would be appropriate for OP to respond to your request for clarification. Although think the first comment does it for me.
    $endgroup$
    – RRL
    Jan 24 at 6:33












  • $begingroup$
    @RRL, can you give a proof of your claim i.e. $F'(0) = f(0) = 0$?
    $endgroup$
    – Tom.
    Jan 24 at 8:19










  • $begingroup$
    @Tom: I added the proof of that. Note that functions which are derivatives on an interval $[a,b]$ are said to belong to the space $mathcal{D}^1([a,b])$ and I am sure (aside from the counterexample I came up with) that $f in mathcal{D}^1([a,b])$ does not imply $|f| in mathcal{D}^1([a,b])$, but I have not found a reference yet.
    $endgroup$
    – RRL
    Jan 24 at 8:51








  • 1




    $begingroup$
    @Tom: If $f$ is Riemann integrable then $g = |f|$ is always Riemann integrable. In my example $f$ is Riemann integrable. I don’t know what you mean by “I assumed”. In general if $g$ is a derivative it does not have to be integrable. Look up Volterra’s function.
    $endgroup$
    – RRL
    Jan 24 at 15:09


















  • $begingroup$
    Well, I'm replying to your comment here. Yes, I agree with you. All I was saying is that the OP hadn't worded the question as precisely as possible. Anyway, +1 for your answer.
    $endgroup$
    – stressed out
    Jan 24 at 6:31






  • 1




    $begingroup$
    @stressedout: Thank you. It would be appropriate for OP to respond to your request for clarification. Although think the first comment does it for me.
    $endgroup$
    – RRL
    Jan 24 at 6:33












  • $begingroup$
    @RRL, can you give a proof of your claim i.e. $F'(0) = f(0) = 0$?
    $endgroup$
    – Tom.
    Jan 24 at 8:19










  • $begingroup$
    @Tom: I added the proof of that. Note that functions which are derivatives on an interval $[a,b]$ are said to belong to the space $mathcal{D}^1([a,b])$ and I am sure (aside from the counterexample I came up with) that $f in mathcal{D}^1([a,b])$ does not imply $|f| in mathcal{D}^1([a,b])$, but I have not found a reference yet.
    $endgroup$
    – RRL
    Jan 24 at 8:51








  • 1




    $begingroup$
    @Tom: If $f$ is Riemann integrable then $g = |f|$ is always Riemann integrable. In my example $f$ is Riemann integrable. I don’t know what you mean by “I assumed”. In general if $g$ is a derivative it does not have to be integrable. Look up Volterra’s function.
    $endgroup$
    – RRL
    Jan 24 at 15:09
















$begingroup$
Well, I'm replying to your comment here. Yes, I agree with you. All I was saying is that the OP hadn't worded the question as precisely as possible. Anyway, +1 for your answer.
$endgroup$
– stressed out
Jan 24 at 6:31




$begingroup$
Well, I'm replying to your comment here. Yes, I agree with you. All I was saying is that the OP hadn't worded the question as precisely as possible. Anyway, +1 for your answer.
$endgroup$
– stressed out
Jan 24 at 6:31




1




1




$begingroup$
@stressedout: Thank you. It would be appropriate for OP to respond to your request for clarification. Although think the first comment does it for me.
$endgroup$
– RRL
Jan 24 at 6:33






$begingroup$
@stressedout: Thank you. It would be appropriate for OP to respond to your request for clarification. Although think the first comment does it for me.
$endgroup$
– RRL
Jan 24 at 6:33














$begingroup$
@RRL, can you give a proof of your claim i.e. $F'(0) = f(0) = 0$?
$endgroup$
– Tom.
Jan 24 at 8:19




$begingroup$
@RRL, can you give a proof of your claim i.e. $F'(0) = f(0) = 0$?
$endgroup$
– Tom.
Jan 24 at 8:19












$begingroup$
@Tom: I added the proof of that. Note that functions which are derivatives on an interval $[a,b]$ are said to belong to the space $mathcal{D}^1([a,b])$ and I am sure (aside from the counterexample I came up with) that $f in mathcal{D}^1([a,b])$ does not imply $|f| in mathcal{D}^1([a,b])$, but I have not found a reference yet.
$endgroup$
– RRL
Jan 24 at 8:51






$begingroup$
@Tom: I added the proof of that. Note that functions which are derivatives on an interval $[a,b]$ are said to belong to the space $mathcal{D}^1([a,b])$ and I am sure (aside from the counterexample I came up with) that $f in mathcal{D}^1([a,b])$ does not imply $|f| in mathcal{D}^1([a,b])$, but I have not found a reference yet.
$endgroup$
– RRL
Jan 24 at 8:51






1




1




$begingroup$
@Tom: If $f$ is Riemann integrable then $g = |f|$ is always Riemann integrable. In my example $f$ is Riemann integrable. I don’t know what you mean by “I assumed”. In general if $g$ is a derivative it does not have to be integrable. Look up Volterra’s function.
$endgroup$
– RRL
Jan 24 at 15:09




$begingroup$
@Tom: If $f$ is Riemann integrable then $g = |f|$ is always Riemann integrable. In my example $f$ is Riemann integrable. I don’t know what you mean by “I assumed”. In general if $g$ is a derivative it does not have to be integrable. Look up Volterra’s function.
$endgroup$
– RRL
Jan 24 at 15:09











0












$begingroup$

Consider $f(x)=x$ which is differentiable everywhere but $|f(x)|$ is not differentiable at zero.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Did you answer the question?
    $endgroup$
    – Jacky Chong
    Jan 24 at 4:48










  • $begingroup$
    It's not your fault. The question is ambiguous. You haven't actually answered the question. Check the comments.
    $endgroup$
    – stressed out
    Jan 24 at 4:50






  • 1




    $begingroup$
    @stressedout: The question is not ambiguous. This answer is about differentiability of $|f|$ -- which does not answer the OP. OP is asking if $f = F'$ on some domain does that mean that $|f| = G'$ for some differentiable function $G$.
    $endgroup$
    – RRL
    Jan 24 at 5:50










  • $begingroup$
    @RRL The question is ambiguous in the sense that the OP has not defined what a 'derivative' function is. I left a comment under their post asking for more explanation because it sounded ambiguous to me at first. Plus, if $f$ is not Riemann integrable, then I don't see how $f=F'$ can hold in the first place. Because if it does, then $int f = f$ exists. No?
    $endgroup$
    – stressed out
    Jan 24 at 5:55












  • $begingroup$
    @stressedout: Good you asked for clarification.
    $endgroup$
    – RRL
    Jan 24 at 6:02
















0












$begingroup$

Consider $f(x)=x$ which is differentiable everywhere but $|f(x)|$ is not differentiable at zero.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Did you answer the question?
    $endgroup$
    – Jacky Chong
    Jan 24 at 4:48










  • $begingroup$
    It's not your fault. The question is ambiguous. You haven't actually answered the question. Check the comments.
    $endgroup$
    – stressed out
    Jan 24 at 4:50






  • 1




    $begingroup$
    @stressedout: The question is not ambiguous. This answer is about differentiability of $|f|$ -- which does not answer the OP. OP is asking if $f = F'$ on some domain does that mean that $|f| = G'$ for some differentiable function $G$.
    $endgroup$
    – RRL
    Jan 24 at 5:50










  • $begingroup$
    @RRL The question is ambiguous in the sense that the OP has not defined what a 'derivative' function is. I left a comment under their post asking for more explanation because it sounded ambiguous to me at first. Plus, if $f$ is not Riemann integrable, then I don't see how $f=F'$ can hold in the first place. Because if it does, then $int f = f$ exists. No?
    $endgroup$
    – stressed out
    Jan 24 at 5:55












  • $begingroup$
    @stressedout: Good you asked for clarification.
    $endgroup$
    – RRL
    Jan 24 at 6:02














0












0








0





$begingroup$

Consider $f(x)=x$ which is differentiable everywhere but $|f(x)|$ is not differentiable at zero.






share|cite|improve this answer









$endgroup$



Consider $f(x)=x$ which is differentiable everywhere but $|f(x)|$ is not differentiable at zero.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 4:48









CyclotomicFieldCyclotomicField

2,4431314




2,4431314








  • 1




    $begingroup$
    Did you answer the question?
    $endgroup$
    – Jacky Chong
    Jan 24 at 4:48










  • $begingroup$
    It's not your fault. The question is ambiguous. You haven't actually answered the question. Check the comments.
    $endgroup$
    – stressed out
    Jan 24 at 4:50






  • 1




    $begingroup$
    @stressedout: The question is not ambiguous. This answer is about differentiability of $|f|$ -- which does not answer the OP. OP is asking if $f = F'$ on some domain does that mean that $|f| = G'$ for some differentiable function $G$.
    $endgroup$
    – RRL
    Jan 24 at 5:50










  • $begingroup$
    @RRL The question is ambiguous in the sense that the OP has not defined what a 'derivative' function is. I left a comment under their post asking for more explanation because it sounded ambiguous to me at first. Plus, if $f$ is not Riemann integrable, then I don't see how $f=F'$ can hold in the first place. Because if it does, then $int f = f$ exists. No?
    $endgroup$
    – stressed out
    Jan 24 at 5:55












  • $begingroup$
    @stressedout: Good you asked for clarification.
    $endgroup$
    – RRL
    Jan 24 at 6:02














  • 1




    $begingroup$
    Did you answer the question?
    $endgroup$
    – Jacky Chong
    Jan 24 at 4:48










  • $begingroup$
    It's not your fault. The question is ambiguous. You haven't actually answered the question. Check the comments.
    $endgroup$
    – stressed out
    Jan 24 at 4:50






  • 1




    $begingroup$
    @stressedout: The question is not ambiguous. This answer is about differentiability of $|f|$ -- which does not answer the OP. OP is asking if $f = F'$ on some domain does that mean that $|f| = G'$ for some differentiable function $G$.
    $endgroup$
    – RRL
    Jan 24 at 5:50










  • $begingroup$
    @RRL The question is ambiguous in the sense that the OP has not defined what a 'derivative' function is. I left a comment under their post asking for more explanation because it sounded ambiguous to me at first. Plus, if $f$ is not Riemann integrable, then I don't see how $f=F'$ can hold in the first place. Because if it does, then $int f = f$ exists. No?
    $endgroup$
    – stressed out
    Jan 24 at 5:55












  • $begingroup$
    @stressedout: Good you asked for clarification.
    $endgroup$
    – RRL
    Jan 24 at 6:02








1




1




$begingroup$
Did you answer the question?
$endgroup$
– Jacky Chong
Jan 24 at 4:48




$begingroup$
Did you answer the question?
$endgroup$
– Jacky Chong
Jan 24 at 4:48












$begingroup$
It's not your fault. The question is ambiguous. You haven't actually answered the question. Check the comments.
$endgroup$
– stressed out
Jan 24 at 4:50




$begingroup$
It's not your fault. The question is ambiguous. You haven't actually answered the question. Check the comments.
$endgroup$
– stressed out
Jan 24 at 4:50




1




1




$begingroup$
@stressedout: The question is not ambiguous. This answer is about differentiability of $|f|$ -- which does not answer the OP. OP is asking if $f = F'$ on some domain does that mean that $|f| = G'$ for some differentiable function $G$.
$endgroup$
– RRL
Jan 24 at 5:50




$begingroup$
@stressedout: The question is not ambiguous. This answer is about differentiability of $|f|$ -- which does not answer the OP. OP is asking if $f = F'$ on some domain does that mean that $|f| = G'$ for some differentiable function $G$.
$endgroup$
– RRL
Jan 24 at 5:50












$begingroup$
@RRL The question is ambiguous in the sense that the OP has not defined what a 'derivative' function is. I left a comment under their post asking for more explanation because it sounded ambiguous to me at first. Plus, if $f$ is not Riemann integrable, then I don't see how $f=F'$ can hold in the first place. Because if it does, then $int f = f$ exists. No?
$endgroup$
– stressed out
Jan 24 at 5:55






$begingroup$
@RRL The question is ambiguous in the sense that the OP has not defined what a 'derivative' function is. I left a comment under their post asking for more explanation because it sounded ambiguous to me at first. Plus, if $f$ is not Riemann integrable, then I don't see how $f=F'$ can hold in the first place. Because if it does, then $int f = f$ exists. No?
$endgroup$
– stressed out
Jan 24 at 5:55














$begingroup$
@stressedout: Good you asked for clarification.
$endgroup$
– RRL
Jan 24 at 6:02




$begingroup$
@stressedout: Good you asked for clarification.
$endgroup$
– RRL
Jan 24 at 6:02


















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