设z=f(x,y)在R^2上是连续的,且limf(x,y)=+无穷,在x^2+y^2=+无穷,证明f(x,y)有最小

Question

设z=f(x,y)在R^2上是连续的,且limf(x,y)=+无穷,在x^2+y^2=+无穷,证明f(x,y)有最小

Answer 1

设A = f(0,0).
由lim{x²+y²→+∞} f(x,y) = +∞,存在r > 0,使x²+y² > r²时恒有f(x,y) > A.
设D为以(0,0)为圆心,r为半径的闭圆盘,即D = {(x,y)|x²+y² ≤ r²}.
由D为有界闭集,即D为紧集,又f在D上连续,可知f在D上可取得最小值.
设(a,b) ∈ D满足f(a,b) ≤ f(x,y)对任意(x,y) ∈ D成立.
则f(a,b) ≤ f(0,0) = A < f(x,y)对任意(x,y) ∈ R²-D也成立.
因此f(a,b)就是f(x,y)在R²上的最小值.