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.