Partial derivatives (question I am grading).

[MathematicalPhysicist]

We have a function f:R^2->R and it has partial derivative of 2nd order.
Show that f_{xy}=0 \forall (x,y)\in \mathbb{R}^2 \Leftrightarrow f(x,y)=g(x)+h(y)

The <= is self explanatory, the => I am not sure I got the right reasoning.

I mean we know that from the above we have: f_x=F(x) (it's a question before this one), but now besides taking an integral I don't see how to show the consequent.

Any thoughts how to show this without invoking integration?

 

[hunt_mat]

My first thought is the mean value theorem which states that if f&#039;(x)=0 then f\equivconstant., apply this to x when y yields \partial_{y}f\equivconstant, but this constant will be dependent on y and therefor is a function of y.

Get the general idea now?

 

[MathematicalPhysicist]

That's basically what I have written, so I know that f_x =F(x) and f_y=G(y). But this is where I am not sure how to procceed.

I mean I know that: if F(x)=h'(x), then G(x,y)= f(x,y)-h(x) then G_x=0 and then G=g(y).

The problem is how do I know that F(x)=h'(x) I am assuming that F is of this form, aren't I?

 

[hunt_mat]

I think that as you effectively have the equation f&#039;(x)=g(x) then you want to show that there is a function F(x) with the property F&#039;(x)=f(x), and I think that this follows from a version of the fundamental theorem of calculus. By extension G(x)=F(x)+C will also satisfy this equation. I think you can say this because f\in C^{1} as you have f&#039;(x).