I need to prove that a function $f:[0,1]\times[0,1]\rightarrow [0,1]$, which is nondecreasing in each variable, is grounded (i.e. that $f(0,y)=0=f(x,0)$ for all $(x,y)$ in $[0,1]\times[0,1]$). Additionally, it is known that $f(t,1)=t=f(1,t)$ for all $t$ in $[0,1]$.
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Since $f$ is nondecreasing in each variable, $f(t,1) \geq f(t,y) \; \textrm{for} \; y \leq 1$. Since the range of $f \in [0,1]$, $0 = f(0,1) \geq f(0,y) \geq 0 \Rightarrow f(0,y) = 0.$
Similarly for the $x$ variable.
user1462620
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Because $f(t, 1) = t = f(1 , t)$ for all $t$ in $[0, 1]$, that means that $f(0, 1) = 0 = f(1, 0)$. This also implies that $f$ is grounded, which follows straightforward from the def'ns.
Don Larynx
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How do you show that $f(0,1)=0=f(1,0)\Rightarrow f(0,y)=0=f(x,0)$ for all $(x,y)$? – user109107 Nov 17 '13 at 03:43
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Additionally, it is known that f(t,1)=t=f(1,t) for all t in [0,1]. – Don Larynx Nov 17 '13 at 03:44
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Let t= 0 @user109107. – Don Larynx Nov 17 '13 at 03:45
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If I let $t=0$, that only proves $f(0,1)=0=f(1,0)$. It does not follow from that alone that $f(0,y)=0=f(x,0)$ for all $(x,y)$. For example, if $f(x,y)=x+y-1$, then $f(0,1/2)=-1/2$. – user109107 Nov 17 '13 at 03:54