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Supposing $f: [0,\infty) \to [0,\infty)$. The goal is to make an increasing function from $f$ using the following rule:-

If $t_1 \leq t_2$ and $f(t_1) > f(t_2)$ then change the value of $f(t_1)$ to $f(t_2)$.

After this change, we have $f(t_1) = f(t_2)$.

Let $g$ be the function resulting from applying the above rule for all $t_1,t_2$ recursively (recursively because if $f(t_2)$ changes then the value of $f(t_1)$ needs to be re-computed)

Is it correct to treat $g$ as a well defined (increasing) function?

Thanks, Phanindra

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You can't do recurssion here, because $[0,\infty)$ is uncountable. You could try $g(x) = \inf\limits_{y\geq x} f(y)$ –  Stefan Dec 21 '12 at 11:23
    
When transforming f into an increasing functions, which properties of f are to be preserved? –  mkl Dec 21 '12 at 11:37
    
Stefan: Thanks for the nice observation (that recursion is not possible). $g(x) = \inf\limits_{y\geq x} f(y)$ works perfectly for me. –  jpv Dec 21 '12 at 14:20
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1 Answer

up vote 3 down vote accepted

Consider the function $f(x)=1/x$ if $x>0$, with also $f(0)=0$. Then for example $f(1)$ will, for any $n>1$, get changed to $n$ on considering that $f(1/n)=n>f(1)=1$. Once this is done there will still be plenty of other $m>1$ for which $f(1/m)=m$ where $m>n$, so that $f(1)$ will have to be changed again from its present value of $n$ to the new value $m>n$. In this way, for this example, there will not be a finite value for $f(1)$ as the process is iterated, and the resulting function will not be defined at $x=1$.

EDIT: As mkl points out in a comment, the interpretion in the above example has the construction backward. When $f(a)>f(b)$ where $a<b$ the jvp construction is to replace $f(a)$ by the "later value" $f(b)$. In this version there is no problem with infinite values occuring, as a value of $f(x)$ is only decreased during the construction, and the decreasing is bounded below by $0$ because the original $f$ is nonnegative. In fact, if $g(x)$ denotes the constructed function, and if we interpret the "iterative procedure" in a reasonable way, it seems one has $$g(x)=\inf \{f(t):t \ge x \},$$ which is a nondecreasing function for any given $f(x)$. Note that Stefan made exactly this suggestion.

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Actually @jpv replaced the value of the function at the lower x $t_1$ to the value at the higher one $t_2$. The problem essentially is the same, though. –  mkl Dec 21 '12 at 12:13
    
coffeemath: Thanks for the answer. The infimum definition was the one I was after. –  jpv Dec 21 '12 at 14:24
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