Putnam 2009, Problem A-6

Let $f : [0, 1]^2 \to \mathbb{R}$ be a continuous function on the closed unit square such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist and are continuous on the interior $(0, 1)^2$. Let $a = \int_0^1 f(0, y) \, dy$, $b = \int_0^1 f(1, y) \, dy$, $c = \int_0^1 f(x, 0) \, dx$, $d = \int_0^1 f(x, 1) \, dx$. Prove or disprove: There must be a point $(x_0, y_0)$ in $(0, 1)^2$ such that
\[
\frac{\partial f}{\partial x}(x_0, y_0) = b - a \quad \text{and} \quad \frac{\partial f}{\partial y}(x_0, y_0) = d - c.
\]

Solution:
We disprove the assertion using the example
\[
f(x,y) = 3(1 + y)(2x - 1)^2 - y.
\]
We have \( b - a = d - c = 0 \) because the identity \( f(x,y) = f(1 - x,y) \) forces \( a = b \), and because
\[
c = \int_0^1 3(2x - 1)^2 dx = 1, \quad d = \int_0^1 (6(2x - 1)^2 - 1) dx = 1.
\]

**Remark.** This problem amounts to refuting a potential generalization of the Mean Value Theorem to bivariate functions. Many counterexamples are possible. Kent Merryfield suggests \( y\sin(2\pi x) \), for which all four of the boundary integrals vanish; here the partial derivatives are \( 2\pi y\cos(2\pi x) \) and \( \sin(2\pi x) \). Catalin Zara suggests \( x^{1/3}y^{2/3} \). Qingchun Ren suggests \( xy(1 - y) \).