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

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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.