Given a function $f(u) \le 2{\sqrt u}\,f({\sqrt u}) + 1$. We need to bound the growth of $f(u)$.
If we expand this by recursively substituting, we get some series like
$1 + \left\{ 2u^{\frac{1}{2}},\hspace{5 pt} 2^2u^{\frac{1}{2}+{\frac{1}{4}}},\hspace{5 pt} 2^3u^{\frac{1}{2}+{\frac{1}{4}}+\frac{1}{8}},\hspace{5 pt} \dots\hspace{5 pt}\right\}$
what would be the ending term? How do we find the bound (in terms of Big O notation ..like $O(\log u)$ etc) ?
$u^{\frac12 + \frac14 + \frac 18 + \cdots} \rightarrow u$ for all $u$ and $2^n \rightarrow \infty$ so the bound is infinite if $u > 0$ and $1+0$ if $u = 0$.