2009 Putnam A6

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.

$$

令 $f:[0,1]^2 \to \mathbb{R}$ 为闭单元上的连续函数

平方使得 $\frac{\partial f}{\partial x}$ 和 $\frac{\partial f}{\partial y}$ 存在

且在内部 $(0,1)^2$ 上连续。设 $a = \int_0^1 f(0,y)\,dy$,

$b = \int_0^1 f(1,y)\,dy$，$c = \int_0^1 f(x,0)\,dx$，$d = \int_0^1 f(x,1)\,dx$。

证明或反驳：$(0,1)^2$ 中必须存在一个点 $(x_0,y_0)$ 使得

$$

\frac{\partial f}{\partial x} (x_0,y_0) = b - a

\quad \text{和} \quad

\frac{\partial f}{\partial y} (x_0,y_0) = d - c。

$$