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
    2 days ago










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










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






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    2 days 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
    2 days ago


















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
    2 days ago










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










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






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    2 days 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
    2 days ago
















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 2 days ago









fazanfazan

608




608












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










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










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






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    2 days 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
    2 days ago




















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










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










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






  • 1




    $begingroup$
    @avs That is false.
    $endgroup$
    – zhw.
    2 days 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
    2 days ago


















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




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












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




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












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




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




1




1




$begingroup$
@avs That is false.
$endgroup$
– zhw.
2 days ago




$begingroup$
@avs That is false.
$endgroup$
– zhw.
2 days 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
2 days 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
2 days 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
    2 days ago










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






  • 1




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










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










    • $begingroup$
      Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
      $endgroup$
      – Robert Furber
      2 days 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 2 days ago

























    answered 2 days ago









    Haris GusicHaris Gusic

    3,546627




    3,546627












    • $begingroup$
      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
      2 days ago










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










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












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




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






    • 1




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






    • 1




      $begingroup$
      @HarisGusic yes I realized as I posted. Fixed it
      $endgroup$
      – K.Power
      2 days 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 2 days ago

























    answered 2 days ago









    K.PowerK.Power

    3,709926




    3,709926












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






    • 1




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


















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






    • 1




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
















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




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




    1




    1




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




    $begingroup$
    @HarisGusic yes I realized as I posted. Fixed it
    $endgroup$
    – K.Power
    2 days 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 2 days ago









        avs

        4,203515




        4,203515










        answered 2 days ago









        ManRowManRow

        25618




        25618






























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            body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分
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            connect to host localhost port 22: Connection refused

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 1 1 I created a Ubuntu Vagrant box and ssh'd into it. Now, created a ssh key and trying to ssh to my localhost but I get the following error. vagrant@vagrant-ubuntu-trusty-64:~$ ssh -vvvv localhost OpenSSH_6.6.1, OpenSSL 1.0.1f 6 Jan 2014 debug1: Reading configuration data /etc/ssh/ssh_config debug1: /etc/ssh/ssh_config line 19: Applying options for * debug2: ssh_connect: needpriv 0 debug1: Connecting to localhost [127.0.0.1] port 22. debug1: connect to address 127.0.0.1 port 22: Connection refused ssh: connect to host localhost port 22: Connection refused When I check the list of ports opened as follows: vagrant@vagrant-ubuntu-trusty-64:~$ s
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            Getting a Wifi WPA2 wifi connection

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 0 I've been trying to get a simple wifi connection ! bu I have to admit I just can't get it done ! here is what I've tried: root@MYONE:/etc# clear root@MYONE:/etc# vi wpa_supplicant.conf ctrl_interface=/var/run/wpa_supplicant ctrl_interface_group=0 update_config=1 network={ ssid="Mi Engine" psk="7407352688e4" } root@MYONE:/etc# iw wlan0 link Not connected. root@MYONE:/etc# ifconfig eth0 Link encap:Ethernet HWaddr 74:FE:48:3A:55:AB inet addr:192.168.178.88 Bcast:192.168.178.255 Mask:255.255.255.0 inet6 addr: fe80::76fe:48ff:fe3a:55ab%4804152/64 Scope:Link UP BRO
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