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The implication of zero mixed partial derivatives for multivariate function's minimization

If the second mixed partial is identically zero, then the function can be written as a sum $f(x,y) = f_1(x) + f_2(y)$$dfracpartial^2 fpartial x partial y = 0 nRightarrow f(x,y) = g(x) + h(y)$If a nonnegative function of $x_1,dots,x_n$ can be written as $sum g_k(x_k)$, then the summands can be taken nonnegative$f(x,y) = f_1(x) + f_2(x)$ with continuous differential real-valued functionsShow that both mixed partial derivatives exist at the origin but are not equalFunction on $mathbbR^2-0$.Let $F_1,F_2:bf R^2 to R$ be functions such that…How many n-th Order Partial Derivatives Exist for a Function of k Variables?Maximize $f(textbfx,textbfy) = f_1(textbfx) + f_2(textbfx,textbfy)$Symmetry of higher order mixed partial derivatives under weaker assumptionsWhy does the partial of $f: Delta to mathbbR^2$ fail to exist?To find function satisfying given partial derivatives2 variable, 2 valued function $f(x_1,,x_2)=(x_1,,x_2)$What will be $frac partial F_i partial x_j$?

2

$begingroup$

Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_12=f''_21=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_textbf x f(textbf x)equiv min_x_1f_1(x_1)+ min_x_2f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$

A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.

multivariable-calculus optimization partial-derivative

share|cite|improve this question

edited Nov 1 '15 at 20:04

user147263

asked Jun 24 '13 at 22:43

jorter.jijorter.ji

258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
  $endgroup$
  – user147263
  Dec 29 '15 at 0:24
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_12=f''_21=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_textbf x f(textbf x)equiv min_x_1f_1(x_1)+ min_x_2f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$

A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.

multivariable-calculus optimization partial-derivative

share|cite|improve this question

edited Nov 1 '15 at 20:04

user147263

asked Jun 24 '13 at 22:43

jorter.jijorter.ji

258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
  $endgroup$
  – user147263
  Dec 29 '15 at 0:24
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_12=f''_21=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_textbf x f(textbf x)equiv min_x_1f_1(x_1)+ min_x_2f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$

A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.

multivariable-calculus optimization partial-derivative

share|cite|improve this question

edited Nov 1 '15 at 20:04

user147263

asked Jun 24 '13 at 22:43

jorter.jijorter.ji

258

$endgroup$

share|cite|improve this question

edited Nov 1 '15 at 20:04

user147263

asked Jun 24 '13 at 22:43

jorter.jijorter.ji

258

share|cite|improve this question

edited Nov 1 '15 at 20:04

user147263

asked Jun 24 '13 at 22:43

jorter.jijorter.ji

258

asked Jun 24 '13 at 22:43

jorter.jijorter.ji

258

asked Jun 24 '13 at 22:43

jorter.jijorter.ji

258

2 Answers
2

active

oldest

votes

1

$begingroup$

For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_xy ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_yx$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.

share|cite|improve this answer

answered Jun 24 '13 at 23:14

Shuhao CaoShuhao Cao

16.1k34292

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 6:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
  $endgroup$
  – Shuhao Cao
  Jun 26 '13 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
  $endgroup$
  – Shuhao Cao
  Jun 26 '13 at 16:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 17:41
  

  
  
  
  

 | 
show 5 more comments

0

$begingroup$

The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.

Here I try to provide a proof without that assumption.

Restatement

I can restate the conjecture with little weaker conditions:

If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.

Proof

From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.

$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$

Annotation

The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.

share|cite|improve this answer

answered Mar 16 at 8:52

TA123TA123

958

$endgroup$

add a comment | 

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1

$begingroup$

For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_xy ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_yx$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.

share|cite|improve this answer

answered Jun 24 '13 at 23:14

Shuhao CaoShuhao Cao

16.1k34292

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 6:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
  $endgroup$
  – Shuhao Cao
  Jun 26 '13 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
  $endgroup$
  – Shuhao Cao
  Jun 26 '13 at 16:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 17:41
  

  
  
  
  

 | 
show 5 more comments

1

$begingroup$

For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_xy ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_yx$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.

share|cite|improve this answer

answered Jun 24 '13 at 23:14

Shuhao CaoShuhao Cao

16.1k34292

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 6:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
  $endgroup$
  – Shuhao Cao
  Jun 26 '13 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
  $endgroup$
  – Shuhao Cao
  Jun 26 '13 at 16:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
  $endgroup$
  – jorter.ji
  Jun 26 '13 at 17:41
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_xy ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_yx$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.

share|cite|improve this answer

answered Jun 24 '13 at 23:14

Shuhao CaoShuhao Cao

16.1k34292

$endgroup$

share|cite|improve this answer

answered Jun 24 '13 at 23:14

Shuhao CaoShuhao Cao

16.1k34292

answered Jun 24 '13 at 23:14

Shuhao CaoShuhao Cao

16.1k34292

answered Jun 24 '13 at 23:14

Shuhao CaoShuhao Cao

16.1k34292

0

$begingroup$

The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.

Here I try to provide a proof without that assumption.

Restatement

I can restate the conjecture with little weaker conditions:

If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.

Proof

From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.

$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$

Annotation

The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.

share|cite|improve this answer

answered Mar 16 at 8:52

TA123TA123

958

$endgroup$

add a comment | 

0

$begingroup$

The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.

Here I try to provide a proof without that assumption.

Restatement

I can restate the conjecture with little weaker conditions:

If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.

Proof

From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.

$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$

Annotation

The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.

share|cite|improve this answer

answered Mar 16 at 8:52

TA123TA123

958

$endgroup$

add a comment | 

$begingroup$

The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.

Here I try to provide a proof without that assumption.

Restatement

I can restate the conjecture with little weaker conditions:

If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.

Proof

From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.

$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$

Annotation

The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.

share|cite|improve this answer

answered Mar 16 at 8:52

TA123TA123

958

$endgroup$

share|cite|improve this answer

answered Mar 16 at 8:52

TA123TA123

958

answered Mar 16 at 8:52

TA123TA123

958

answered Mar 16 at 8:52

TA123TA123

958