Confusion about non-derivable continuous functions The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are 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(?) $f(fracx+y2)=fracf(x)+f(y)2$Confused About Limit Points and Closed SetsConfusion About Differentiability of Function
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Confusion about non-derivable continuous functions
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are 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(?) $f(fracx+y2)=fracf(x)+f(y)2$Confused About Limit Points and Closed SetsConfusion About Differentiability of Function
$begingroup$
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
asked Apr 9 at 20:47
fazanfazan
608
$endgroup$
add a comment |
$begingroup$
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
asked Apr 9 at 20:47
fazanfazan
608
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Apr 9 at 20:47
fazanfazan
608
$endgroup$
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
asked Apr 9 at 20:47
fazanfazan
608
asked Apr 9 at 20:47
fazanfazan
608
asked Apr 9 at 20:47
fazanfazan
608
asked Apr 9 at 20:47
fazanfazan
608
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$f(x,y)=begincases(x^2+y^2)sin(frac1sqrtx^2+y^2) &(x,y)neq(0,0)\0&(x,y)=(0,0)endcases$$
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.
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbbRtomathbb R$ (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
$$fleft(xright)=left{beginmatrix1, xinmathbbQ\0,xnotinmathbb Qendmatrixright.$$
edited Apr 10 at 4:11
avs
4,212515
answered Apr 9 at 23:28
ManRowManRow
25618
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
$endgroup$
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.
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
edited Apr 9 at 20:56
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,569627
3,569627
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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
$$f(x,y)=begincases(x^2+y^2)sin(frac1sqrtx^2+y^2) &(x,y)neq(0,0)\0&(x,y)=(0,0)endcases$$
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.
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$f(x,y)=begincases(x^2+y^2)sin(frac1sqrtx^2+y^2) &(x,y)neq(0,0)\0&(x,y)=(0,0)endcases$$
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.
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$f(x,y)=begincases(x^2+y^2)sin(frac1sqrtx^2+y^2) &(x,y)neq(0,0)\0&(x,y)=(0,0)endcases$$
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.
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
$endgroup$
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
$$f(x,y)=begincases(x^2+y^2)sin(frac1sqrtx^2+y^2) &(x,y)neq(0,0)\0&(x,y)=(0,0)endcases$$
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.
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
edited Apr 9 at 21:18
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
answered Apr 9 at 21:10
K.PowerK.Power
3,734926
3,734926
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbbRtomathbb R$ (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
$$fleft(xright)=left{beginmatrix1, xinmathbbQ\0,xnotinmathbb Qendmatrixright.$$
edited Apr 10 at 4:11
avs
4,212515
answered Apr 9 at 23:28
ManRowManRow
25618
$endgroup$
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbbRtomathbb R$ (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
$$fleft(xright)=left{beginmatrix1, xinmathbbQ\0,xnotinmathbb Qendmatrixright.$$
edited Apr 10 at 4:11
avs
4,212515
answered Apr 9 at 23:28
ManRowManRow
25618
$endgroup$
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbbRtomathbb R$ (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
$$fleft(xright)=left{beginmatrix1, xinmathbbQ\0,xnotinmathbb Qendmatrixright.$$
edited Apr 10 at 4:11
avs
4,212515
answered Apr 9 at 23:28
ManRowManRow
25618
$endgroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbbRtomathbb R$ (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
$$fleft(xright)=left{beginmatrix1, xinmathbbQ\0,xnotinmathbb Qendmatrixright.$$
edited Apr 10 at 4:11
avs
4,212515
answered Apr 9 at 23:28
ManRowManRow
25618
edited Apr 10 at 4:11
avs
4,212515
edited Apr 10 at 4:11
avs
4,212515
edited Apr 10 at 4:11
avs
4,212515
4,212515
answered Apr 9 at 23:28
ManRowManRow
25618
answered Apr 9 at 23:28
ManRowManRow
25618
answered Apr 9 at 23:28
ManRowManRow
25618
25618
add a comment |
add a comment |
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For $|x|$ its derivative isn't continuous t zero.
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– coffeemath
Apr 9 at 20:51
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Where did you read that erroneous definition?
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– bof
Apr 9 at 21:01
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lecture notes by my prof. i might be mosreading them though
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– fazan
Apr 9 at 21:04
1
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@avs That is false.
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– zhw.
Apr 9 at 22:07
2
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@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).
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– Robert Furber
Apr 10 at 0:19