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

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.

• 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. – user147263 Dec 29 '15 at 0:24

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.

• Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants. – jorter.ji Jun 26 '13 at 6:13
• @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. – Shuhao Cao Jun 26 '13 at 12:32
• Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$? – jorter.ji Jun 26 '13 at 16:49
• @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. – Shuhao Cao Jun 26 '13 at 16:52
• 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. – jorter.ji Jun 26 '13 at 17:41

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.