How Do We Show Integral Unity of Transformed Functions?

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SUMMARY

The discussion centers on demonstrating the integral unity of the transformed function g(y) derived from a given function f on R, where \(\int_R f(x)dx = 1\), f is decreasing, and 1-Lipschitz. The transformation is defined as g(y) = min{x,f(x)} with y = x - f(x) for x ≥ 0. The participants explore the necessary conditions for maintaining integral unity, referencing the work of B.F. Logan and L.A. Shepp on variational problems and the application of Lebesgue measure in solving such problems.

PREREQUISITES
  • Understanding of Lebesgue measure theory
  • Familiarity with Lipschitz continuity
  • Knowledge of integral calculus
  • Basic concepts of topological transformations
NEXT STEPS
  • Study Lebesgue integration techniques
  • Explore Lipschitz functions and their properties
  • Investigate topological transformations in analysis
  • Review variational problems in mathematical literature
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Mathematicians, analysts, and students in advanced calculus or real analysis who are interested in integral transformations and their properties.

hnh
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Hello, I tried this in analysis but maybe it is a more topological question. If given a function f on R such that \int_R f(x)dx = 1 and is decreasing and 1-lipschitz, show that
the function g(y) = min{x,f(x)} where y = x-f(x) and x>=0, also satisfies \int_Y g(y)dy=1.

I really would appreciate any observations. The transformation of the Domain is 1-1 right? so what else
is required to have integral unity of g also? I have some other info that may or may not help: f(x) = g(y) -1/2y +1/2|y|, x=g(y)+1/2y +1/2|y|.
Thank you
 
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hnh said:
Hello, I tried this in analysis but maybe it is a more topological question. If given a function f on R such that \int_R f(x)dx = 1 and is decreasing and 1-lipschitz

I doubt that such a function exists.
 
I refer you to the bibliography item in my thesis:

1. B.F. Logan and L.A. Shepp. A Variational Problem for Random Young Tableaux. Ad-
vances in Mathematics, 26, 1977, 206-222.

The problem is solved now using lebesgue measure...
 

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