I need a check with the following exercise:
Let $w: [0, \infty) \rightarrow [0,\infty)$ with $w(0)=0$ and $w(y)>0$ a continuous function. Let $y:[0,\infty) \rightarrow [0,\infty)$ a $C^1$ function with the property
\begin{cases} |y'(x)| \leq w(y(x)) \\ y(0)=0 \end{cases}
for all $x \geq0$
Show that if $\int_0^1 \frac{dy}{w(y)}=+\infty$, then $y(x)=0$ for all $x \geq0$
Here's my attempt:
From the Cauchy problem I find, for all $x\geq0$: $$-w(y(x))\leq y'(x) \leq w(y(x))$$
Let $ c \geq 0$: from the rightmost part: $$\int_0^c\frac{y'(x)}{w(y(x))}dx \leq c$$
which is equivalent to (setting $\phi=y(x)$) $$\int_0^{y(c)}\frac{d \phi}{w(\phi)} \leq c$$
If $y(c)>1$, then I obtain $$\int_0^{1}\frac{d \phi}{w(\phi)} + \int_1^{y(c)}\frac{d \phi}{w(\phi)} \leq c$$ By assumption, the first term is $+\infty$, which implies a contradiction ($\infty \leq c$). Therefore it has to be $y(c) < 1$.
From the other condition, I obtain:
$$\int_0^{y(c)}\frac{d \phi}{w(\phi)} \geq-c (<0)$$
which is always satisfied. I don't know how to continue in order to show that $y$ has to be $0$.