Confusion about non-derivable continuous functions












2












$begingroup$


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
    9 hours ago










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    9 hours ago










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    9 hours ago






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    7 hours ago






  • 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
    5 hours ago


















2












$begingroup$


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
    9 hours ago










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    9 hours ago










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    9 hours ago






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    7 hours ago






  • 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
    5 hours ago
















2












2








2


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 9 hours ago









fazanfazan

537




537












  • $begingroup$
    For $|x|$ its derivative isn't continuous t zero.
    $endgroup$
    – coffeemath
    9 hours ago










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    9 hours ago










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    9 hours ago






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    7 hours ago






  • 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
    5 hours ago




















  • $begingroup$
    For $|x|$ its derivative isn't continuous t zero.
    $endgroup$
    – coffeemath
    9 hours ago










  • $begingroup$
    Where did you read that erroneous definition?
    $endgroup$
    – bof
    9 hours ago










  • $begingroup$
    lecture notes by my prof. i might be mosreading them though
    $endgroup$
    – fazan
    9 hours ago






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    7 hours ago






  • 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
    5 hours ago


















$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
9 hours ago




$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
9 hours ago












$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
9 hours ago




$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
9 hours ago












$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
9 hours ago




$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
9 hours ago




1




1




$begingroup$
@avs That is false.
$endgroup$
– zhw.
7 hours ago




$begingroup$
@avs That is false.
$endgroup$
– zhw.
7 hours ago




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
5 hours ago






$begingroup$
@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
5 hours ago












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
    5 hours ago










  • $begingroup$
    Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
    $endgroup$
    – Robert Furber
    5 hours ago



















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
    8 hours ago






  • 1




    $begingroup$
    @HarisGusic yes I realized as I posted. Fixed it
    $endgroup$
    – K.Power
    8 hours ago



















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
      5 hours ago










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      5 hours ago
















    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
      5 hours ago










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      5 hours ago














    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 9 hours ago

























    answered 9 hours ago









    Haris GusicHaris Gusic

    3,516627




    3,516627












    • $begingroup$
      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
      5 hours ago










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      5 hours ago


















    • $begingroup$
      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
      5 hours ago










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      5 hours ago
















    $begingroup$
    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
    5 hours ago




    $begingroup$
    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
    5 hours ago












    $begingroup$
    Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
    $endgroup$
    – Robert Furber
    5 hours ago




    $begingroup$
    Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
    $endgroup$
    – Robert Furber
    5 hours ago











    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
      8 hours ago






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      8 hours ago
















    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
      8 hours ago






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      8 hours ago














    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 8 hours ago

























    answered 8 hours ago









    K.PowerK.Power

    3,710926




    3,710926












    • $begingroup$
      The partial derivatives need not be continuous for differentiability.
      $endgroup$
      – Haris Gusic
      8 hours ago






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      8 hours ago


















    • $begingroup$
      The partial derivatives need not be continuous for differentiability.
      $endgroup$
      – Haris Gusic
      8 hours ago






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      8 hours ago
















    $begingroup$
    The partial derivatives need not be continuous for differentiability.
    $endgroup$
    – Haris Gusic
    8 hours ago




    $begingroup$
    The partial derivatives need not be continuous for differentiability.
    $endgroup$
    – Haris Gusic
    8 hours ago




    1




    1




    $begingroup$
    @HarisGusic yes I realized as I posted. Fixed it
    $endgroup$
    – K.Power
    8 hours ago




    $begingroup$
    @HarisGusic yes I realized as I posted. Fixed it
    $endgroup$
    – K.Power
    8 hours ago











    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 1 hour ago









        avs

        4,012515




        4,012515










        answered 6 hours ago









        ManRowManRow

        25618




        25618






























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            Verónica Boquete

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            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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