Let F and y both be continuous for simplicity. Knowing that:[tex]

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Discussion Overview

The discussion revolves around the implications of the integral equation involving the functions F and y, specifically whether the function y can be considered bounded given the relationship defined by the integral. The context includes mathematical reasoning and exploration of the properties of continuous functions.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant questions whether the function y is bounded based on the integral equation involving F and y.
  • Another participant suggests that taking derivatives leads to the conclusion that |y(x)| = 1 for all x > 0, implying a specific boundedness.
  • A later reply introduces a condition where if F(x) = C, then the integral condition leads to 0 = C, indicating that if F(x) is identically zero, no conclusions can be drawn about y.
  • One participant acknowledges a previous oversight in their reasoning regarding the boundedness of y.

Areas of Agreement / Disagreement

Participants express differing views on the boundedness of y, with some asserting that it is bounded under certain conditions while others highlight exceptions that prevent a definitive conclusion.

Contextual Notes

The discussion does not resolve the implications of the integral condition fully, particularly in cases where F(x) may be constant or zero, leaving some assumptions unexamined.

Malmstrom
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Let F and y both be continuous for simplicity. Knowing that:
[tex]\int_0^x F'(t)y^2(t) dt = F(x) \quad \forall x \geq 0[/tex]
can you say that the function [tex]y[/tex] is bounded? Why? I know that [tex]\int_0^x F'(t) dt = F(x)[/tex] but I can't find a suitable inequality to prove rigorously that y is bounded.
 
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Malmstrom said:
Let F and y both be continuous for simplicity. Knowing that:
[tex]\int_0^x F'(t)y^2(t) dt = F(x) \quad \forall x \geq 0[/tex]
can you say that the function [tex]y[/tex] is bounded? Why? I know that [tex]\int_0^x F'(t) dt = F(x)[/tex] but I can't find a suitable inequality to prove rigorously that y is bounded.

Taking derivatives of both sides F'(x)y2(x)=F'(x) for all x>0, so |y(x)|=1.
 


mathman said:
Taking derivatives of both sides F'(x)y2(x)=F'(x) for all x>0, so |y(x)|=1.

Unless, [itex]F(x) = C[/itex] in which case the condition you gave us is [itex]0 = C[/itex]. So, if [itex]F(x) \equiv 0[/itex] we can't say anything about the function [itex]y(x)[/itex].
 


mathman said:
Taking derivatives of both sides F'(x)y2(x)=F'(x) for all x>0, so |y(x)|=1.

Thanks, I was missing something very easy.
 

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