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.
 A: 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.
A: 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.
