How to verify that a function is $C^1$ (continuously differentiable)? Possible connection with component-wise...
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Let $f:mathbb{R}^n to mathbb{R}^m$ be a function. We write $f=left(f_1,cdots,f_mright)$, where $f_i:mathbb{R}^n to mathbb{R}$.
Original Question : The problem from which I was motivated to ask this question is the following : Let $f:mathbb{R}^2 to mathbb{R}^2$ be given by $fleft(x,yright)=left(cos{x}+cos{y},,sin{x}+sin{y}right)$. Is it true that $f in C^1left(mathbb{R}^2right)?$
The answer is obviously yes! But to establish that, if I understand it correctly, I have to do a hell lot of work. I have to compute jacobian matrix, check continuity of the entries, establish differentiability of $f$ and then check continuity of the total derivative. I was wondering if there's any easy formulation to verify $C^1$-ness.
Question : I want to know whether or not the following statement holds in general : $f in C^1left(mathbb{R}^nright)$ if and only if $f_i in C^1left(mathbb{R}^nright)$ for every $i=1,cdots,n$. If yes, why? If not, is there any such (sufficient) condition at all?
real-analysis analysis multivariable-calculus derivatives continuity
asked Nov 18 at 4:13
Dragon
568114
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0
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Let $f:mathbb{R}^n to mathbb{R}^m$ be a function. We write $f=left(f_1,cdots,f_mright)$, where $f_i:mathbb{R}^n to mathbb{R}$.
Original Question : The problem from which I was motivated to ask this question is the following : Let $f:mathbb{R}^2 to mathbb{R}^2$ be given by $fleft(x,yright)=left(cos{x}+cos{y},,sin{x}+sin{y}right)$. Is it true that $f in C^1left(mathbb{R}^2right)?$
The answer is obviously yes! But to establish that, if I understand it correctly, I have to do a hell lot of work. I have to compute jacobian matrix, check continuity of the entries, establish differentiability of $f$ and then check continuity of the total derivative. I was wondering if there's any easy formulation to verify $C^1$-ness.
Question : I want to know whether or not the following statement holds in general : $f in C^1left(mathbb{R}^nright)$ if and only if $f_i in C^1left(mathbb{R}^nright)$ for every $i=1,cdots,n$. If yes, why? If not, is there any such (sufficient) condition at all?
real-analysis analysis multivariable-calculus derivatives continuity
asked Nov 18 at 4:13
Dragon
568114
Yes, your equivalence is true. This follows quite easily from the definition (the Jacobian of $f$ is the Jacobians of $f_1,dots, f_m$ stacked on top of each other).
– PhoemueX
Nov 18 at 8:52
@PhoemueX Thanks for the comment! So, can I assume that we only need the entries of the Jacobian $frac{partial f_i}{partial x_j}$ to be $C^1$? (It's probably a very basic question, but I need a confirmation) For example, in the case of the function $f:mathbb{R}^2 to mathbb{R}^2$ mentioned above, I get a nice Jacobian with nice $4$ entries which are all $C^1$. Is it enough to say that $f$ is $C^1$?
– Dragon
Nov 18 at 9:33
Yes, all you need is that all partial derivatives $partial f_i /partial x_j$ exist and are continuous. That this suffices is one of the theorems of multivariable calculus.
– PhoemueX
Nov 18 at 15:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:mathbb{R}^n to mathbb{R}^m$ be a function. We write $f=left(f_1,cdots,f_mright)$, where $f_i:mathbb{R}^n to mathbb{R}$.
Original Question : The problem from which I was motivated to ask this question is the following : Let $f:mathbb{R}^2 to mathbb{R}^2$ be given by $fleft(x,yright)=left(cos{x}+cos{y},,sin{x}+sin{y}right)$. Is it true that $f in C^1left(mathbb{R}^2right)?$
The answer is obviously yes! But to establish that, if I understand it correctly, I have to do a hell lot of work. I have to compute jacobian matrix, check continuity of the entries, establish differentiability of $f$ and then check continuity of the total derivative. I was wondering if there's any easy formulation to verify $C^1$-ness.
Question : I want to know whether or not the following statement holds in general : $f in C^1left(mathbb{R}^nright)$ if and only if $f_i in C^1left(mathbb{R}^nright)$ for every $i=1,cdots,n$. If yes, why? If not, is there any such (sufficient) condition at all?
real-analysis analysis multivariable-calculus derivatives continuity
asked Nov 18 at 4:13
Dragon
568114
Let $f:mathbb{R}^n to mathbb{R}^m$ be a function. We write $f=left(f_1,cdots,f_mright)$, where $f_i:mathbb{R}^n to mathbb{R}$.
Original Question : The problem from which I was motivated to ask this question is the following : Let $f:mathbb{R}^2 to mathbb{R}^2$ be given by $fleft(x,yright)=left(cos{x}+cos{y},,sin{x}+sin{y}right)$. Is it true that $f in C^1left(mathbb{R}^2right)?$
The answer is obviously yes! But to establish that, if I understand it correctly, I have to do a hell lot of work. I have to compute jacobian matrix, check continuity of the entries, establish differentiability of $f$ and then check continuity of the total derivative. I was wondering if there's any easy formulation to verify $C^1$-ness.
Question : I want to know whether or not the following statement holds in general : $f in C^1left(mathbb{R}^nright)$ if and only if $f_i in C^1left(mathbb{R}^nright)$ for every $i=1,cdots,n$. If yes, why? If not, is there any such (sufficient) condition at all?
real-analysis analysis multivariable-calculus derivatives continuity
real-analysis analysis multivariable-calculus derivatives continuity
asked Nov 18 at 4:13
Dragon
568114
asked Nov 18 at 4:13
Dragon
568114
asked Nov 18 at 4:13
Dragon
568114
asked Nov 18 at 4:13
Dragon
568114
asked Nov 18 at 4:13
Dragon
568114
568114
Yes, your equivalence is true. This follows quite easily from the definition (the Jacobian of $f$ is the Jacobians of $f_1,dots, f_m$ stacked on top of each other).
– PhoemueX
Nov 18 at 8:52
@PhoemueX Thanks for the comment! So, can I assume that we only need the entries of the Jacobian $frac{partial f_i}{partial x_j}$ to be $C^1$? (It's probably a very basic question, but I need a confirmation) For example, in the case of the function $f:mathbb{R}^2 to mathbb{R}^2$ mentioned above, I get a nice Jacobian with nice $4$ entries which are all $C^1$. Is it enough to say that $f$ is $C^1$?
– Dragon
Nov 18 at 9:33
Yes, all you need is that all partial derivatives $partial f_i /partial x_j$ exist and are continuous. That this suffices is one of the theorems of multivariable calculus.
– PhoemueX
Nov 18 at 15:38
add a comment |
Yes, your equivalence is true. This follows quite easily from the definition (the Jacobian of $f$ is the Jacobians of $f_1,dots, f_m$ stacked on top of each other).
– PhoemueX
Nov 18 at 8:52
@PhoemueX Thanks for the comment! So, can I assume that we only need the entries of the Jacobian $frac{partial f_i}{partial x_j}$ to be $C^1$? (It's probably a very basic question, but I need a confirmation) For example, in the case of the function $f:mathbb{R}^2 to mathbb{R}^2$ mentioned above, I get a nice Jacobian with nice $4$ entries which are all $C^1$. Is it enough to say that $f$ is $C^1$?
– Dragon
Nov 18 at 9:33
Yes, all you need is that all partial derivatives $partial f_i /partial x_j$ exist and are continuous. That this suffices is one of the theorems of multivariable calculus.
– PhoemueX
Nov 18 at 15:38
Yes, your equivalence is true. This follows quite easily from the definition (the Jacobian of $f$ is the Jacobians of $f_1,dots, f_m$ stacked on top of each other).
– PhoemueX
Nov 18 at 8:52
Yes, your equivalence is true. This follows quite easily from the definition (the Jacobian of $f$ is the Jacobians of $f_1,dots, f_m$ stacked on top of each other).
– PhoemueX
Nov 18 at 8:52
@PhoemueX Thanks for the comment! So, can I assume that we only need the entries of the Jacobian $frac{partial f_i}{partial x_j}$ to be $C^1$? (It's probably a very basic question, but I need a confirmation) For example, in the case of the function $f:mathbb{R}^2 to mathbb{R}^2$ mentioned above, I get a nice Jacobian with nice $4$ entries which are all $C^1$. Is it enough to say that $f$ is $C^1$?
– Dragon
Nov 18 at 9:33
@PhoemueX Thanks for the comment! So, can I assume that we only need the entries of the Jacobian $frac{partial f_i}{partial x_j}$ to be $C^1$? (It's probably a very basic question, but I need a confirmation) For example, in the case of the function $f:mathbb{R}^2 to mathbb{R}^2$ mentioned above, I get a nice Jacobian with nice $4$ entries which are all $C^1$. Is it enough to say that $f$ is $C^1$?
– Dragon
Nov 18 at 9:33
Yes, all you need is that all partial derivatives $partial f_i /partial x_j$ exist and are continuous. That this suffices is one of the theorems of multivariable calculus.
– PhoemueX
Nov 18 at 15:38
Yes, all you need is that all partial derivatives $partial f_i /partial x_j$ exist and are continuous. That this suffices is one of the theorems of multivariable calculus.
– PhoemueX
Nov 18 at 15:38
add a comment |
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Yes, your equivalence is true. This follows quite easily from the definition (the Jacobian of $f$ is the Jacobians of $f_1,dots, f_m$ stacked on top of each other).
– PhoemueX
Nov 18 at 8:52
@PhoemueX Thanks for the comment! So, can I assume that we only need the entries of the Jacobian $frac{partial f_i}{partial x_j}$ to be $C^1$? (It's probably a very basic question, but I need a confirmation) For example, in the case of the function $f:mathbb{R}^2 to mathbb{R}^2$ mentioned above, I get a nice Jacobian with nice $4$ entries which are all $C^1$. Is it enough to say that $f$ is $C^1$?
– Dragon
Nov 18 at 9:33
Yes, all you need is that all partial derivatives $partial f_i /partial x_j$ exist and are continuous. That this suffices is one of the theorems of multivariable calculus.
– PhoemueX
Nov 18 at 15:38