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(?) $f(fracx+y2)=fracf(x)+f(y)2$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 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(?) $f(fracx+y2)=fracf(x)+f(y)2$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 2 days ago
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 2 days ago
fazanfazan
608
$endgroup$
$begingroup$
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
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 2 days ago
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 2 days ago
fazanfazan
608
asked 2 days ago
fazanfazan
608
asked 2 days ago
fazanfazan
608
asked 2 days ago
fazanfazan
608
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
add a comment |
$begingroup$
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
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 2 days ago
Haris GusicHaris Gusic
3,546627
$endgroup$
$begingroup$
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
yesterday
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 2 days ago
K.PowerK.Power
3,709926
$endgroup$
$begingroup$
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
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 yesterday
avs
4,203515
answered 2 days ago
ManRowManRow
25618
$endgroup$
add a comment |
Your Answer
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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 2 days ago
Haris GusicHaris Gusic
3,546627
$endgroup$
$begingroup$
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
yesterday
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 2 days ago
Haris GusicHaris Gusic
3,546627
$endgroup$
$begingroup$
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
yesterday
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 2 days ago
Haris GusicHaris Gusic
3,546627
$endgroup$
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 2 days ago
Haris GusicHaris Gusic
3,546627
edited 2 days ago
answered 2 days ago
Haris GusicHaris Gusic
3,546627
answered 2 days ago
Haris GusicHaris Gusic
3,546627
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
yesterday
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
2 days ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
$begingroup$
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
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
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 2 days ago
K.PowerK.Power
3,709926
$endgroup$
$begingroup$
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
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 2 days ago
K.PowerK.Power
3,709926
$endgroup$
$begingroup$
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
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 2 days ago
K.PowerK.Power
3,709926
$endgroup$
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 2 days ago
K.PowerK.Power
3,709926
edited 2 days ago
answered 2 days ago
K.PowerK.Power
3,709926
answered 2 days ago
K.PowerK.Power
3,709926
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
add a comment |
$begingroup$
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
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 yesterday
avs
4,203515
answered 2 days ago
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 yesterday
avs
4,203515
answered 2 days ago
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 yesterday
avs
4,203515
answered 2 days ago
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 yesterday
avs
4,203515
answered 2 days ago
ManRowManRow
25618
edited yesterday
avs
4,203515
edited yesterday
avs
4,203515
edited yesterday
avs
4,203515
4,203515
answered 2 days ago
ManRowManRow
25618
answered 2 days ago
ManRowManRow
25618
answered 2 days ago
ManRowManRow
25618
25618
add a comment |
add a comment |
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-continuity, derivatives, functions, real-analysis
$begingroup$
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
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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
2 days ago