Confusion about non-derivable continuous functions












3












$begingroup$


I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.



And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.



I am almost certain I am getting something wrong here, but I can not even pin-point what.






real-analysis functions derivatives continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    For $|x|$ its derivative isn't continuous t zero.
    $endgroup$
    – coffeemath
    Apr 9 at 20:51










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    Apr 9 at 21:01










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    Apr 9 at 21:04






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    Apr 9 at 22:07






  • 2




    $begingroup$
    @avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
    $endgroup$
    – Robert Furber
    Apr 10 at 0:19


















3












$begingroup$


I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.



And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.



I am almost certain I am getting something wrong here, but I can not even pin-point what.






real-analysis functions derivatives continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    For $|x|$ its derivative isn't continuous t zero.
    $endgroup$
    – coffeemath
    Apr 9 at 20:51










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    Apr 9 at 21:01










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    Apr 9 at 21:04






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    Apr 9 at 22:07






  • 2




    $begingroup$
    @avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
    $endgroup$
    – Robert Furber
    Apr 10 at 0:19
















3












3








3


1



$begingroup$


I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.



And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.



I am almost certain I am getting something wrong here, but I can not even pin-point what.






real-analysis functions derivatives continuity







share|cite|improve this question









$endgroup$




I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.



And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.



I am almost certain I am getting something wrong here, but I can not even pin-point what.






real-analysis functions derivatives continuity




real-analysis functions derivatives continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 9 at 20:47









fazanfazan

608




608












  • $begingroup$
    For $|x|$ its derivative isn't continuous t zero.
    $endgroup$
    – coffeemath
    Apr 9 at 20:51










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    Apr 9 at 21:01










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    Apr 9 at 21:04






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    Apr 9 at 22:07






  • 2




    $begingroup$
    @avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
    $endgroup$
    – Robert Furber
    Apr 10 at 0:19




















  • $begingroup$
    For $|x|$ its derivative isn't continuous t zero.
    $endgroup$
    – coffeemath
    Apr 9 at 20:51










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    Apr 9 at 21:01










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    Apr 9 at 21:04






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    Apr 9 at 22:07






  • 2




    $begingroup$
    @avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
    $endgroup$
    – Robert Furber
    Apr 10 at 0:19


















$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
Apr 9 at 20:51




$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
Apr 9 at 20:51












$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
Apr 9 at 21:01




$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
Apr 9 at 21:01












$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
Apr 9 at 21:04




$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
Apr 9 at 21:04




1




1




$begingroup$
@avs That is false.
$endgroup$
– zhw.
Apr 9 at 22:07




$begingroup$
@avs That is false.
$endgroup$
– zhw.
Apr 9 at 22:07




2




2




$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
Apr 10 at 0:19






$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
Apr 10 at 0:19












3 Answers
3






active

oldest

votes


















4












$begingroup$

That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.



The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
    $endgroup$
    – Robert Furber
    Apr 10 at 0:20










  • $begingroup$
    Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
    $endgroup$
    – Robert Furber
    Apr 10 at 0:24



















1












$begingroup$

As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.



If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The partial derivatives need not be continuous for differentiability.
    $endgroup$
    – Haris Gusic
    Apr 9 at 21:13






  • 1




    $begingroup$
    @HarisGusic yes I realized as I posted. Fixed it
    $endgroup$
    – K.Power
    Apr 9 at 21:13



















1












$begingroup$

Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).



And finally, note that some functions can even be nowhere-continuous as well! Such as
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$






share|cite|improve this answer











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.



    The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
      $endgroup$
      – Robert Furber
      Apr 10 at 0:20










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      Apr 10 at 0:24
















    4












    $begingroup$

    That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.



    The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
      $endgroup$
      – Robert Furber
      Apr 10 at 0:20










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      Apr 10 at 0:24














    4












    4








    4





    $begingroup$

    That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.



    The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.






    share|cite|improve this answer











    $endgroup$



    That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.



    The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 9 at 20:56

























    answered Apr 9 at 20:51









    Haris GusicHaris Gusic

    3,570627




    3,570627












    • $begingroup$
      It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
      $endgroup$
      – Robert Furber
      Apr 10 at 0:20










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      Apr 10 at 0:24


















    • $begingroup$
      It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
      $endgroup$
      – Robert Furber
      Apr 10 at 0:20










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      Apr 10 at 0:24
















    $begingroup$
    It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
    $endgroup$
    – Robert Furber
    Apr 10 at 0:20




    $begingroup$
    It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
    $endgroup$
    – Robert Furber
    Apr 10 at 0:20












    $begingroup$
    Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
    $endgroup$
    – Robert Furber
    Apr 10 at 0:24




    $begingroup$
    Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
    $endgroup$
    – Robert Furber
    Apr 10 at 0:24











    1












    $begingroup$

    As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.



    If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
    $$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
    at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The partial derivatives need not be continuous for differentiability.
      $endgroup$
      – Haris Gusic
      Apr 9 at 21:13






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      Apr 9 at 21:13
















    1












    $begingroup$

    As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.



    If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
    $$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
    at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The partial derivatives need not be continuous for differentiability.
      $endgroup$
      – Haris Gusic
      Apr 9 at 21:13






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      Apr 9 at 21:13














    1












    1








    1





    $begingroup$

    As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.



    If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
    $$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
    at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.






    share|cite|improve this answer











    $endgroup$



    As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.



    If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
    $$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
    at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 9 at 21:18

























    answered Apr 9 at 21:10









    K.PowerK.Power

    3,744926




    3,744926












    • $begingroup$
      The partial derivatives need not be continuous for differentiability.
      $endgroup$
      – Haris Gusic
      Apr 9 at 21:13






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      Apr 9 at 21:13


















    • $begingroup$
      The partial derivatives need not be continuous for differentiability.
      $endgroup$
      – Haris Gusic
      Apr 9 at 21:13






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      Apr 9 at 21:13
















    $begingroup$
    The partial derivatives need not be continuous for differentiability.
    $endgroup$
    – Haris Gusic
    Apr 9 at 21:13




    $begingroup$
    The partial derivatives need not be continuous for differentiability.
    $endgroup$
    – Haris Gusic
    Apr 9 at 21:13




    1




    1




    $begingroup$
    @HarisGusic yes I realized as I posted. Fixed it
    $endgroup$
    – K.Power
    Apr 9 at 21:13




    $begingroup$
    @HarisGusic yes I realized as I posted. Fixed it
    $endgroup$
    – K.Power
    Apr 9 at 21:13











    1












    $begingroup$

    Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).



    And finally, note that some functions can even be nowhere-continuous as well! Such as
    $$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).



      And finally, note that some functions can even be nowhere-continuous as well! Such as
      $$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).



        And finally, note that some functions can even be nowhere-continuous as well! Such as
        $$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$






        share|cite|improve this answer











        $endgroup$



        Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).



        And finally, note that some functions can even be nowhere-continuous as well! Such as
        $$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 10 at 4:11









        avs

        4,197515




        4,197515










        answered Apr 9 at 23:28









        ManRowManRow

        25618




        25618






























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            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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