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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 $\min_{\textbf x} f(\textbf x)\equiv \min_{x_1}f_1(x_1)+ \min_{x_2}f_2(x_2)$? 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.

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

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  • $\begingroup$ 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
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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.

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