证明 给定$(x_0,y_0)\in U$，由于$U$是开集，故存在$(x_0,y_0)$的邻域$V\subseteq U$。由$f(x,y)$对$x$的连续性可得$\mathrm{\forall }\epsilon >0$，$\mathrm{\exists }\delta (\epsilon ,x_0,y_0)>0$，使得$\mathrm{\forall }(x,y)\in V$，都有$$\begin{array}{}\text{(1)}& |x-x_0|<\delta \Rightarrow |f(x,y_0)-f(x_0,y_0)|<\frac{\epsilon }{2}\end{array}$$ 由$f(x,y)$的偏导数有界可得$|\frac{\partial f}{\partial y}(x,y)|\le M$（$M>0$）对所有$(x,y)\in U$成立。由Lagrange中值定理可得$$\begin{array}{}\text{(2)}& f(x,y)-f(x,y_0)=\frac{\partial f}{\partial y}(x,\xi )(y-y_0)\Rightarrow |f(x,y)-f(x,y_0)|\le M|y-y_0|\end{array}$$ 取${\delta }^{\prime }=min\{\delta ,\frac{\epsilon }{2M}\}>0$，则有$$\begin{array}{}\text{(3)}& \begin{array}{rl}\parallel (x,y)-(x_0,y_0)\parallel <{\delta }^{\prime }& \Rightarrow |x-x_0|<{\delta }^{\prime }<\delta \Rightarrow |f(x,y_0)-f(x_0,y_0)|<\frac{\epsilon }{2}\\ & \Rightarrow |y-y_0|<{\delta }^{\prime }\Rightarrow |f(x,y)-f(x,y_0)|\le {\delta }^{\prime }M<\frac{\epsilon }{2}\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(4)}& |f(x,y)-f(x_0,y_0)|\le |f(x,y)-f(x,y_0)|+|f(x,y_0)-f(x_0,y_0)|<\epsilon \end{array}$$ 由$(x_0,y_0)$的任意性可知$f$在$U$上连续。 $◻$

证明 (1) 设$(x_0,y_0)$为$f$在$\bar{D}$上的最小值点。假设$f(x_0,y_0)<0$，则$(x_0,y_0)\in D$；又$f\in {\mathcal{C}}^2(D)$，由Fermat引理可知$(x_0,y_0)$为驻点，且$0>f(x_0,y_0)=\mathrm{\Delta }f(x_0,y_0)=\mathrm{tr}{H}_f(x_0,y_0)\ge 0$，矛盾！故$f(x_0,y_0)\ge 0$，即$f\ge 0$对所有$(x,y)\in D$成立。

(2) 记$m:=\underset{(x,y)\in \partial D}{min}f(x,y)>0$、$X:=\underset{(x,y)\in \partial D}{max}x<+\mathrm{\infty }$。构造函数$$\begin{array}{}\text{(6)}& g(x,y):=f(x,y)-m{\mathrm{e}}^{x-X},(x,y)\in D\end{array}$$ 计算可得$$\begin{array}{}\text{(7)}& \begin{array}{rl}\mathrm{\Delta }g& =\mathrm{\Delta }f-m\mathrm{\Delta }{\mathrm{e}}^{x-X}=f-m{\mathrm{e}}^{x-X}=g,(x,y)\in D\\ g& =f-m{\mathrm{e}}^{x-X}\ge f-m\ge 0,(x,y)\in \partial D\end{array}\end{array}$$ 由(1)可得$g\ge 0$对所有$(x,y)\in D$成立，因此$$\begin{array}{}\text{(8)}& f(x,y)=g(x,y)+m{\mathrm{e}}^{x-X}\ge m{\mathrm{e}}^{x-X}>0,(x,y)\in D\end{array}$$ 故$f>0$对所有$(x,y)\in D$成立。 $◻$