I'm seeking a better proof of the following fact: If $g$ is a non-negative bounded function, $g(0)=0$ and $g'(t)\leq \sqrt{g(t)}$ for all $t>0$, then $g(t)\leq t^2/4$.
The upper bound $t^2/4$ is naturally obtained by pretending we have an equality and solving the ODE. A rigorous proof can be obtained by iterating $M$ times to get \begin{align} g(t)&\leq \int_0^t\sqrt{\int_0^{t_2}\sqrt{\int_0^{t_3}\cdots\sqrt{\int_0^{t_{M}}g(t_{M+1})}}}\\ &\leq \sqrt[2M]{\|g\|_{\infty}}\cdot\int_0^t\sqrt{\int_0^{t_2}\sqrt{\int_0^{t_3}\cdots\sqrt{\int_0^{t_{M}}1}}}. \end{align} Let $M\to\infty$, the first term $\sqrt[2M]{\|g\|_{\infty}}\to 1$ and the second tends to $t^2/4$.
Can we alternatively prove it in a cleaner way, without iteration?
A first attempt is to put $f=\sqrt{g}$. Then we get $f'(t)\leq 1/2$ on $\{g> 0\}$. It is not clear to me how to make it rigorous from here.
Let $f_\epsilon(t)=\sqrt{g(t)+\epsilon^2}$ for $\epsilon>0$. Then $f_\epsilon(0)=\epsilon$ and $2f_\epsilon f'_\epsilon=g'\leq \sqrt{g}=\sqrt{f_\epsilon^2-\epsilon^2}\leq f_\epsilon$. Thus, $f'_\epsilon\leq 1/2$ and $f_\epsilon(t)\leq \epsilon + t/2$ by integration. It follows that $g(t)= f_\epsilon^2(t)-\epsilon^2\leq \epsilon t + t^2/4$ and $g(t)\leq t^2/4$ by letting $\epsilon\to0$.