1
$\begingroup$

For any $\alpha\in [0, 1]$ and $g:[a, b]\to [0, 1]$, let us define $g^{-}:[0, 1]\to [a, b]$, by the formula $g^{-}(\alpha) = \inf (x\in [a, b] ; g(x)\ge \alpha )$.

Let $f_{n}, f : [a, b]\to [0, 1]$, $n\in \mathbb{N}$, be continuous on $[a, b]$, such that the value $1$ is attained, for $f$ and for all $f_{n}$. It is true that if $f_{n}\to f$ uniformly on $[a, b]$, then $f_{n}^{-}(\alpha)\to f^{-}(\alpha)$, almost everywhere in $[0, 1]$ ?

Thank you in advance.

Best, George

Improve this question
$\endgroup$
7
  • $\begingroup$ First you have $g$ defined on $[a,b]$, then you ask a conclusion for any $a \in [0,1]$. Not the same $a$? $\endgroup$
    – Gerald Edgar
    Commented Apr 18, 2012 at 15:37
  • 1
    $\begingroup$ In addition to what Gerald Edgar asks: how do you define the inf in case the set is empty? That being said, as far as I understand the question, I doubt this is true. Say take maps from [0,1] to itself that are straight lines from some value t, say (1-1/n), at 0 to 1 at 1. Let your sequences of functions be such that t converges to 1; this will converge uniformly to constant 1. While this minus transform you consider at 1 should be 1 for all these functions it should be 0 for the limit (if I understand the def correctly). $\endgroup$
    – user9072
    Commented Apr 18, 2012 at 15:55
  • $\begingroup$ One more thought: I guess (but did not check in detail as long as the def is not completely clear) if you change gretear-equal to strict greater this would work. $\endgroup$
    – user9072
    Commented Apr 18, 2012 at 16:20
  • $\begingroup$ I apologize, $a$ in the definitions of $g^{-}$, $f_{n}^{-}$ and $f^{-}$ is in fact $\alpha$ (to be not confused with the endpoint $a$ of the interval $[a, b]$. $\endgroup$
    – George
    Commented Apr 18, 2012 at 16:40
  • $\begingroup$ @quid: it seems to me that if we change the definition of f-minus to make the inequality strict, then it is still seen to be false by a similar counterexample: let f_n be 1/n on the interval [0,1/2] and a straight line on the interval [1/2,1]. For each of these functions f_n, the inf of the x with f_n(x) > 0 is 0, but their limit f is 0 on the interval [0,1/2], so the inf of the x with f(x) > 0 is 1/2. $\endgroup$
    – Trevor Wilson
    Commented Apr 18, 2012 at 17:53

2 Answers 2

Reset to default
1
$\begingroup$

There are two problems:

  1. Replace a continuous $g:[a,b]\to [0,1]$ by its monotone non-decreasing majorant $g^{\text{mon}}$ which is the infimum of all monotone non-increasing functions $\ge g$. I did not check the details, but this operation might be continuous in the $\sup$-norm.

  2. Is taking inverse on the space of monotone non-decreasing functions $f:[a,b]\to [0,1]$ with $f(a)=0$ (?!) and $f(b)=1$ (?!) into the space of monotone non-decreasing functions $[0,1]\to [a,b]$ continuous with respect to the $\sup$-norms?

This two hints might help you to decide the question.

Improve this answer
$\endgroup$
0
$\begingroup$

The inf in case the set is empty is well-defined according to his assumption.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .