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Let $F(v)=\inf\{x: f(x)\leq v\}$, where $f(x)$ is a nonincreasing, right continuous function.

I have already prove that $F(v)$ is nonincreasing also. However, how to prove this is right continuous?

Right coninuous means if $v_m\to v,\ v_m\geq v$, then $F(v_m)\to F(v)$.

This is what I tried:

Pick a sequence {$v_m$} decreasing to $v$. We must have $F(v_m)$ monotone increasing, bounded by $F(v)$ So $F(v_m)$ converge to some number, say $F(\hat v)$. Now we just need to prove $F(\hat v)\geq F(v)$. If this is not the case, i.e.$F(\hat v)<F(v)$. And I am stucked at here.

Any suggestions?

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You may and want to try and use the right-continuity of $f$. –  Eckhard Dec 19 '12 at 21:33
    
Yeah, I see. This is also what I am thinking right now. But I kind of don't know how to use it? –  Zhixia Zhang Dec 19 '12 at 21:34
    
What happens if you compare $F(\nu_m)$ directly with $F(\nu)$? You will naturally have to invoke the right-continuity of $f(x)$. As well, do not assume that convergence is "to some number, say $F(\tilde{\nu})$ (your hope is that $\tilde{v}=v$). Just assume the limit is some number $L$ and show that $L=F(\nu)$ –  Alex R. Dec 19 '12 at 21:35

1 Answer 1

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You know that $\langle F(v_m):m\in\Bbb N\rangle$ converges to some number $y$, but you don’t know that $y$ is in the range of $F$, so you don’t know that it can be written as $F(\hat v)$ for some $\hat v\in\Bbb R$. (Of course you hope to show that $y=F(v)$.)

HINT: Suppose that $y<F(v)=\inf\{x:f(x)\le v\}$; clearly $f(y)>v$, so there is an $m\in\Bbb N$ such that $v<v_m<f(y)$, and hence $y\notin\{x:f(x)\le v_m\}$. Let $A=\{x:f(x)\le v_m\}$. First note that if $u>x\in A$, then $u\in A$, so $y<x$ for all $x\in A$. Now use the right-continuity of $f$ to show that $F(v_m)=\inf A\in A$, and conclude that $y<F(v_m)$, which is a contradiction.

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