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

[Malmstrom]

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

 

[mathman]

Let F and y both be continuous for simplicity. Knowing that:
\int_0^x F'(t)y^2(t) dt = F(x) \quad \forall x \geq 0
can you say that the function y is bounded? Why? I know that \int_0^x F'(t) dt = F(x) 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.

 

[Dickfore]

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

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

 

[Malmstrom]

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.