Prove if $\partial f ^2 / \partial x \partial y= 0$ then there exists $f_1, f_2$ such that $f(x,y) = f_1(x) + f_2(y)$

Question

Show that if $f$ is a twice differentiable, continuously differentiable real-valued function on an open interval in $E^2$ and $\partial f ^2 / \partial x \partial y= 0$ then there are continuously differentiable real-valued functions $f_1, f_2$ on open intervals in $\Bbb R$ such that

$$f(x,y) = f_1(x) + f_2(y)$$

I have an idea of how to prove this but I cannot seem to formulate a rigorous proof. Essentially what I want to say is that $\partial f / \partial y$ is equal to something in terms of only $y$, meaning the $x$ and $y$ terms of $f(x,y)$ are separate.

Answer 1

Notice that $$\frac{\partial h}{\partial x}(x,y)=k(x)\iff h(x,y)=k_1(x)+k_2(y)$$ where $k_1'(x)=k(x)$ since $$\frac{\partial }{\partial x}(k_1(x)+k_2(y))=k_1'(x)=k(x).$$ Moreover $$\frac{\partial h}{\partial x}(x,y)=0\implies h(x,y)=p(y)$$ since $\frac{\partial }{\partial x}p(y)=0.$ Therefore,

$$\frac{\partial^2 f}{\partial x\partial y}=0\implies \frac{\partial f}{\partial x}=f_1(x)\implies f(x,y)=g_1(x)+g_2(y).$$

Answer 2

We can write $\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)= 0$. Doing the integration with respect to x we get $\frac{\partial f}{\partial y}= p(y)$ where p(y) is any function of y. That is, the integral of 0 with respect to x is a "constant" but because the partial derivative treats y as a constant, that "constant" could be any function of y. Then integrating $\frac{\partial f}{\partial y}= p(y)$ with respect to y we get $f= P(y)+ Q(x)$ where P is $\int p(y)dy$ and Q(x) is the "constant of integration" which, again, could be any function of x.