Confusion about non-derivable continuous functions The 2019 Stack Overflow Developer Survey...

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Confusion about non-derivable continuous functions



Confusion about non-derivable continuous functions



The 2019 Stack Overflow Developer Survey Results Are InAre there any implicit, continuous, non-differentiable functions?Logical Relations Between Three Statements about Continuous FunctionsCombination of continuous and discontinuous functionsIs there only one continuous-everywhere non-differentiable function?Intuition behind uniformly continuous functionsWhy weren't continuous functions defined as Darboux functions?Examples of functions that do not belong to any Baire classFind all continuous functions that satisfy the Jensen inequality(?) $f(frac{x+y}{2})=frac{f(x)+f(y)}{2}$Confused About Limit Points and Closed SetsConfusion About Differentiability of Function












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
    2 hours ago










  • $begingroup$
    A function can be continuous at a point without having any derivatives at that point; e.g., coffeemath's example. The statement you probably want is that a function of several variables is differentiable at a point iff all of its partial derivatives exist and are continuous at that point.
    $endgroup$
    – avs
    2 hours ago










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










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










  • $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    59 mins 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
    2 hours ago










  • $begingroup$
    A function can be continuous at a point without having any derivatives at that point; e.g., coffeemath's example. The statement you probably want is that a function of several variables is differentiable at a point iff all of its partial derivatives exist and are continuous at that point.
    $endgroup$
    – avs
    2 hours ago










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










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










  • $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    59 mins 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 2 hours ago









fazanfazan

537




537












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










  • $begingroup$
    A function can be continuous at a point without having any derivatives at that point; e.g., coffeemath's example. The statement you probably want is that a function of several variables is differentiable at a point iff all of its partial derivatives exist and are continuous at that point.
    $endgroup$
    – avs
    2 hours ago










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










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










  • $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    59 mins ago


















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










  • $begingroup$
    A function can be continuous at a point without having any derivatives at that point; e.g., coffeemath's example. The statement you probably want is that a function of several variables is differentiable at a point iff all of its partial derivatives exist and are continuous at that point.
    $endgroup$
    – avs
    2 hours ago










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










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










  • $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    59 mins ago
















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




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












$begingroup$
A function can be continuous at a point without having any derivatives at that point; e.g., coffeemath's example. The statement you probably want is that a function of several variables is differentiable at a point iff all of its partial derivatives exist and are continuous at that point.
$endgroup$
– avs
2 hours ago




$begingroup$
A function can be continuous at a point without having any derivatives at that point; e.g., coffeemath's example. The statement you probably want is that a function of several variables is differentiable at a point iff all of its partial derivatives exist and are continuous at that point.
$endgroup$
– avs
2 hours ago












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




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












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




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












$begingroup$
@avs That is false.
$endgroup$
– zhw.
59 mins ago




$begingroup$
@avs That is false.
$endgroup$
– zhw.
59 mins ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

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











$endgroup$





















    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
      1 hour ago






    • 1




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












    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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











    $endgroup$


















      3












      $begingroup$

      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











      $endgroup$
















        3












        3








        3





        $begingroup$

        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 2 hours ago

























        answered 2 hours ago









        Haris GusicHaris Gusic

        3,506627




        3,506627























            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
              1 hour ago






            • 1




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






            • 1




              $begingroup$
              @HarisGusic yes I realized as I posted. Fixed it
              $endgroup$
              – K.Power
              1 hour 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 1 hour ago

























            answered 1 hour ago









            K.PowerK.Power

            3,710926




            3,710926












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






            • 1




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


















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






            • 1




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
















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




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




            1




            1




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




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


















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