证明函数为连续函数的题怎么做

Question

证明函数为连续函数的题怎么做

Answer 1

可设|f(x,y)|≤1
1.
1>σ>0
A(σ)={(x,y),(x,y)∈[0,1]×[0,1],x≥y+σ},
B(σ)={(x,y),(x,y)∈[0,1]×[0,1],y≥x+σ}.
==>
f(x,y)在A(σ),B(σ)上一致连续.
2.
ⅰ.
x∈[0,1],可设x>0,
任意x/2>ε>0,
f(x,y)在A(ε/8),B(ε/8)上一致连续.
==>
有ε/8>δ>0,当|x-x1|+|y-y1|≤δ时,|f(x,y)-f(x1,y1)|≤ε/4.
==>当0<x-x1≤δ,
ⅱ.
|F(x)-F(x1)|≤∫{0→x-ε/4}|f(x,y)-f(x1,y)|dy+
+∫{x+ε/8→1}|f(x,y)-f(x1,y)|dy+
+∫{x-ε/4→x+ε/8}[|f(x,y)|+|f(x1,y)|]dy=I1+I2+I3
ⅲ.
I3≤2[3ε/8]=3ε/4,
ⅳ.
y≤x-ε/4≤x-ε/8==>(x,y)∈A(ε/8),
y≤x-ε/4=y≤x1-ε/4+x-x1≤x1-ε/4+ε/8=x1-ε/8
==>(x1,y)∈A(ε/8)
而|x-x1|+|y-y|=|x-x1|≤δ
==>
|f(x,y)-f(x1,y)|≤ε/4.
==>
I1≤[ε/4][x-ε/4]
ⅴ.
同理
I2≤[ε/4][1-x-ε/8]
==>
|F(x)-F(x1)|≤I1+I2+I3≤ε
3.
同理
当0<x1-x≤δ,|F(x)-F(x1)|≤ε
==>
F(x)在[0,1]上连续.