Constructing a counter-example of an increasing function such that $|\alpha(x)/x|, |\alpha(x)/\alpha(x/2)| \to \infty$ as $x\to 0^+$

Question

Consider a continuous function $\alpha:[0,\infty) \to [0,\infty)$ satisfying the following properties:

1. $\alpha(0)=0;$
2. $\alpha$ is a strictly increasing function;
3. $\alpha$ is smooth in $(0,\infty);$ and
4. $|\alpha(x)/x| \to \infty$ as $x\to 0^+$.

Under these conditions, is it true that there exists $\delta>0$ and $Μ >0$ such that $$ \left|\frac{\alpha(x)}{\alpha(x/2)}\right| \leq M,\ \forall\ x\in(0,\delta)\ ?$$

I do not know if the above question is either true or false. Could someone please help me to solve this question?

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Searching for a counter-example:

Since I could not prove the above question, I started looking for a counter-example. However, the examples that I was able to find that satisfy 1-4, such as

1. $\alpha(x) = x^\beta$, for $0<\beta<1$;
2. $\alpha(x) = -x \log (x)$ in a neighborhood of $0;$ and
3. $\alpha(x) = (1-x^x)/x$ in a neighborhood of $0.$

However, all these functions fulfil the desired property.

Answer 1

No. Let $\phi:\mathbb R \to [0, 1]$ be a smooth increasing step function, satisfying $\phi(x) = 0$ for $x \le 1$, and $\phi(x) = 1$ for $x \ge 2$. (For example, take the integral of a bump function.)

Then let $a_n = n^{-n}$, and define

$$\alpha(x) = x^{1/2} + \sum_{n=1}^\infty \left(a_n^{1/2} - a_{n+1}^{1/2}\right) \phi\left(\frac{x}{a_{n+1}}\right).$$

You can check that this function is well-defined, and satisfies your conditions, but we have

$$\frac{\alpha(2a_{n+1})}{\alpha(a_{n+1})} \ge \frac{a_n^{1/2}}{2a_{n+1}^{1/2}} \ge \frac{n^{1/2}}2.$$

(As an aside, the result is true if $\alpha(0) = 0$ and $\alpha$ is concave and increasing, which seems like the intuition the conditions are aiming for. This counterexample works because it's not a concave function.)