# 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.

#### Solutions Collecting From Web of "The implication of zero mixed partial derivatives for multivariate function's minimization"

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.