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Please can you help me to find all functions of class $C^2$, $f:\mathbb R^2\to\mathbb R$ such that $\frac{\partial^2f}{\partial x\partial y} = 0$.

Thank you so much!

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yes.. i'm sorry because i have not written correctly – Phương Ami Dec 7 '12 at 8:14
I have made the title more meaningful. – Simon Hayward Dec 7 '12 at 8:30
Here is an outline solution: - Engage your brain. - Solve the problem. – Rhys Dec 7 '12 at 8:38
Note first that $\dfrac{\partial^2{f}}{\partial{x} \partial{y}}=\dfrac{\partial}{\partial{x}}\left(\dfrac{\partial{f}}{\partial{y}} \right).$ Let $g(x, \,y):=\dfrac{\partial{f}}{\partial{y}}.$ What $g(x, \,y)$ satisfies $\dfrac{\partial{g}}{\partial{x}}=0$? – M. Strochyk Dec 7 '12 at 9:09

We will show that a $f\in C^2(\mathbb R^2)$ satisfies $\partial_x\partial_y f = 0$ iff there are $g,h \in C^2(\mathbb R)$ with $f(x,y) = g(x) + h(y)$ for all $x,y \in \mathbb R$.

If we have $\partial_x \partial_y f = 0$, it follows that for $x,y \in \mathbb R$: \begin{align*} \partial_y f(x,y) &= \partial_y f(0,y) + \int_0^x \partial_x\partial_y f(\xi,y) \, d\xi\\ &= \partial_y f(0,y) \end{align*} and hence \begin{align*} f(x,y) &= f(x,0) + \int_0^y \partial_y f(x,\eta)\, d\eta\\ &= f(x,0) + \int_0^y \partial_y f(0, \eta)\, d\eta\\ &= f(x,0) + f(0,y) - f(0,0) \end{align*} So if we let $g(x) := f(x,0)$, $h(y) := f(0,y) - f(0,0)$ we have $g,h \in C^2(\mathbb R)$ and $f = g\pi_1 + h \pi_2$ ($\pi_i \colon \mathbb R^2\to \mathbb R$ denoting the coordinate projections).

On the otherside, if $g,h \in C^2(\mathbb R)$ are such that $f = g\pi_1 + h\pi_2$, we have for $x,y \in \mathbb R$: \begin{align*} \partial_x\partial_y f(x,y) &= \partial_x\partial_y \bigl(g(x) + h(y)\bigr)\\ &= \partial_x h'(y)\\ &= 0. \end{align*}

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