I would like to obtain an explicit description of a continuous, non-decreasing function $f:[0,1]\to [0,\infty )$ with the following properties: (i) $f(0)=0$; (ii) $\int_{0^+} f(x)/x\ dx=\infty$ and (iii) $\liminf_{x\to 0}f(x^2)/f(x)=0$.
The function $f(x)=-1/\ln(x)$ (near zero) initially looks like a good candidate, but fails to satisfy (iii).
Any help appreciated.
Original answer:
There are many such functions; one example is $f(x) = \frac{x}{1-x}$ on $[0, 1)$ -- it is not hard to verify that all three properties hold.
Some motivation: I chose $x$ for the fast decay near $x = 0$, and $\frac{1}{1-x}$ to give a divergent integral around $x = 1$.
Answer to updated question:
Reframe the problem using the substitution $x = e^{-u}$, and define $g(u) = f(e^{-u})$, so our condition on $g : (a, \infty) \to [0, \infty)$ is that it is continuous and nonincreasing with $\lim_{u \to \infty} g(u) = 0$, $\int_b^\infty g(u) \,du = \infty$ for all $b \geq a$, and $$\liminf_{u \to \infty} \frac{g(2u)}{g(u)} = 0.$$ It is not hard to prove that if we strengthen the third condition to $\lim_{u \to \infty} \frac{g(2u)}{g(u)} = 0$, then no such function exists, which suggests that $g$ must have arbitrarily "steep" decreases over some intervals of the form $[u, 2u]$, while overall not decreasing too fast.
For the moment, let's relax the continuity requirement. Then we can try a function of the form $$g(u) = \frac{1}{h(\lfloor h^{-1}(u) \rfloor)}$$ for a very fast-growing positive function $h$, e.g. $h(u) = e^{u^2}$ (here $\lfloor \cdot \rfloor$ denotes the floor function). Intuitively, this function behaves asymptotically like $\frac{1}{u}$, and for $u$ with $$h(n) < u < h(n+1) < 2u < h(n+2),$$ we have $$\frac{g(2u)}{g(u)} = \frac{h(n)}{h(n+1)}$$ which goes to zero as $n \to \infty$, satisfying the second condition.
To check the integral condition, note that $$\int_{h(k)}^\infty g(u) \,du = \sum_{n=k}^\infty \int_{h(n)}^{h(n+1)} g(u) \,du = \sum_{n=k}^\infty \frac{h(n+1) - h(n)}{h(n)} = \infty.$$
Thus this choice of $g$ satisfies all of our requirements except continuity. To make it continuous, we can simply redefine it in a small neighborhood of each discontinuity at $u = h(n)$, so that instead of jumping from $h(n-1)$ to $h(n)$, it undergoes a very rapid linear decrease in that neighborhood. This can then be translated into a function $f$ which satisfies all of the original requirements via $f(x) = g(-\ln x)$.
Unfortunately, this might not be as explicit as you want -- it could be made more explicit if you can find an explicit continuous function which increasingly closely approximates the floor function.