三、（本题共 10 分）设 $D \subset \mathbb{R}^{2}$ 是凸区域，函数 $f(x, y)$ 是凸函数．证明或否定：$f(x, y)$ 在 $D$上连续．

注：函数 $f(x, y)$ 为凸函数的定义是 $\forall \alpha \in(0,1)$ 以及 $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \in D$ ，成立

$$
f\left(\alpha x_{1}+(1-\alpha) x_{2}, \alpha y_{1}+(1-\alpha) y_{2}\right) \leq \alpha f\left(x_{1}, y_{1}\right)+(1-\alpha) f\left(x_{2}, y_{2}\right) .
$$

【证明】：结论成立．分两步证明结论．
（i）对于 $\delta>0$ 以及 $\left[x_{0}-\delta, x_{0}+\delta\right]$ 上的凸函数 $g(x)$ ，容易验证 $\forall x \in\left(x_{0}-\delta, x_{0}+\delta\right)$ ：

$$
\frac{g\left(x_{0}\right)-g\left(x_{0}-\delta\right)}{\delta} \leq \frac{g(x)-g\left(x_{0}\right)}{x-x_{0}} \leq \frac{g\left(x_{0}+\delta\right)-g\left(x_{0}\right)}{\delta}
$$

从而

$$
\left|\frac{g(x)-g\left(x_{0}\right)}{x-x_{0}}\right| \leq\left|\frac{g\left(x_{0}+\delta\right)-g\left(x_{0}\right)}{\delta}\right|+\left|\frac{g\left(x_{0}\right)-g\left(x_{0}-\delta\right)}{\delta}\right|, \forall \in\left(x_{0}-\delta, x_{0}+\delta\right) .
$$

由此即得 $\boldsymbol{g}(\boldsymbol{x})$ 在 $x_{0}$ 连续。一般地，可得开区间上的一元凸函数连续。
（ii）设 $\left(x_{0}, y_{0}\right) \in D$ ，则有 $\delta>0$ 使得

$$
E_{\delta} \equiv\left[x_{0}-\delta, x_{0}+\delta\right] \times\left[y_{0}-\delta, y_{0}+\delta\right] \subset D .
$$

注意到固定 $x$ 或 $y$ 时，$f(x, y)$ 作为一元函数都是凸函数，由（i）的结论， $f\left(x, y_{0}\right), f\left(x, y_{0}+\delta\right), f\left(x, y_{0}-\delta\right)$ 都是 $x \in\left[x_{0}-\delta, x_{0}+\delta\right]$ 上的连续函数，从而它们有界，即存在常数 $M_{\delta}>0$ 使得

$$
\frac{\frac{\left|f\left(x, y_{0}+\delta\right)-f\left(x, y_{0}\right)\right|}{\delta}+\frac{\left|f\left(x, y_{0}\right)-f\left(x, y_{0}-\delta\right)\right|}{\delta}+}{\frac{\left|f\left(x_{0}+\delta, y_{0}\right)-f\left(x_{0}, y_{0}\right)\right|}{\delta}+\frac{\left|f\left(x_{0}, y_{0}\right)-f\left(x_{0}-\delta, y_{0}\right)\right|}{\delta} \leq M_{\delta},}
$$

进一步，由（i）的结论，对于 $(x, y) \in E_{\delta}$ ，

$$
\begin{aligned}
& \left|f(x, y)-f\left(x_{0}, y_{0}\right)\right| \leq\left|f(x, y)-f\left(x, y_{0}\right)\right|+\left|f\left(x, y_{0}\right)-f\left(x_{0}, y_{0}\right)\right| \\
& \leq\left(\frac{\left|f\left(x, y_{0}+\delta\right)-f\left(x, y_{0}\right)\right|}{\delta}+\frac{\left|f\left(x, y_{0}\right)-f\left(x, y_{0}-\delta\right)\right|}{\delta}\right)\left|y-y_{0}\right|+ \\
& \left.\quad \left\lvert\, \frac{\left|f\left(x_{0}+\delta, y_{0}\right)-f\left(x_{0}, y_{0}\right)\right|}{\delta}+\frac{\left|f\left(x_{0}, y_{0}\right)-f\left(x_{0}-\delta, y_{0}\right)\right|}{\delta}\right.\right)\left|x-x_{0}\right| \\
& \quad \leq M_{\delta}\left|y-y_{0}\right|+M_{\delta}\left|x-x_{0}\right| .
\end{aligned}
$$

于是 $f(x, y)$ 在 $\left(x_{0}, y_{0}\right)$ 连续．