Am I using the fundamental theorem of calculus correctly in this proof showing $\partial_x\partial_y f=0$ iff $f(x,y)=g(x)+h(y).$

Question

Let $f: \mathbb R^2\to \mathbb R$ be a $\mathcal C^2$ function. Show that $f$ satisfies $\partial_x\partial_y f=0$ if and only if there exist $\mathcal C^2$ functions, $g$ and $h$, of one variable such that $f(x,y)=g(x)+h(y).$

If $f(x,y)=g(x)+h(y)$, then we get $\partial_x\partial_y f=0$.

If $\partial_x\partial_y f=0$, then $$\int_0^y \partial_x\partial_y f(x,y) \ dx=\int_0^y 0 \ dx$$ $$\implies \partial_yf(x,y)=h_1(y)$$ $$\implies \int_0^x \partial_y f(x,y)\ dy=\int_0^x h_1(y) \ dy$$ $$\implies f(x,y)=h_1(y)g_1(x)+g(x)$$ But since we must have $\partial_y f(x,y)= g(x)$, then $g_1(x)=C$.

Am I using the fundamental theorem of calculus correctly?

Answer 1

Since $\partial_x\partial_y f=0$ we must have $\partial_y f$ independent of $x$ and thus a function of $y$ which you have denoted by $h_1(y)$. This is fine but then you need to integrate with respect to $y$ to get $f(x, y) =h(y) +g(x) $ where $g(x) $ works as constant of integration with respect to $y$ and $h(y) $ is anti-derivative of $h_1(y)$ and we are done.

Answer 2

No, that's not correct. Try setting, for example, $h_1(y)=y$. The answer is not $Ch_1(y)+g(x)$.