Confusion about non-derivable continuous functions
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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
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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
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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).
$endgroup$
– 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
$endgroup$
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
608
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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
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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.
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$begingroup$
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
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
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.
$endgroup$
$begingroup$
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 $f:mathbb{R}tomathbb {R}$ (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
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
$endgroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
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.
edited Apr 9 at 20:56
answered Apr 9 at 20:51
Haris GusicHaris Gusic
3,570627
3,570627
$begingroup$
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
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
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.
$endgroup$
$begingroup$
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
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
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.
$endgroup$
$begingroup$
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
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
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.
$endgroup$
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.
edited Apr 9 at 21:18
answered Apr 9 at 21:10
K.PowerK.Power
3,744926
3,744926
$begingroup$
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 $f:mathbb{R}tomathbb {R}$ (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
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
$endgroup$
add a comment |
$begingroup$
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
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
$endgroup$
add a comment |
$begingroup$
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
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
$endgroup$
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
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
edited Apr 10 at 4:11
avs
4,197515
4,197515
answered Apr 9 at 23:28
ManRowManRow
25618
25618
add a comment |
add a comment |
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$begingroup$
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