# Let F and y both be continuous for simplicity. Knowing that:$$1. Jun 10, 2010 ### Malmstrom Let F and y both be continuous for simplicity. Knowing that: [tex] \int_0^x F'(t)y^2(t) dt = F(x) \quad \forall x \geq 0$$
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.

2. Jun 10, 2010

### mathman

Re: Inequality

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

3. Jun 10, 2010

### Dickfore

Re: Inequality

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 $y(x)$.

4. Jun 10, 2010

### Malmstrom

Re: Inequality

Thanks, I was missing something very easy.

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