Let $f(x)$ satisfies $f(x)\ge|x|^{\alpha}, \frac12\lt\alpha\lt1$ and $f(x)f(2x)\le|x|$ then find $\lim_{x\to0}f(x)$

Question

Let $f(x)$ satisfies $f(x)\ge|x|^{\alpha}, \frac12\lt\alpha\lt1$ and $f(x)f(2x)\le|x|$ for all $x$ in the deleted neighbourhood of zero then $\lim_{x\to0}f(x)=1/\alpha/0/$Does not exist?

$f(x)\ge|x|^\alpha\implies f(2x)\ge|2x|^\alpha$

Multiplying, $f(x)f(2x)\ge2^\alpha|x|^{2\alpha}$

Also, $f(x)f(2x)\le|x|$

Thus, $2^\alpha|x|^{2\alpha}\le|x|$

As an example, I am taking $\alpha=\frac34$

So, $2^\frac34|x|^3\le|x|$

Since $|x|$ is positive and non-zero, so, $2^\frac34|x|^2\le1\implies|x|^2\le2^{-\frac34}\implies|x|\le2^{-\frac38}\implies x\in[-2^{-\frac38}, 2^{-\frac38}]-\{0\}$

I had hoped to comment something about $f(x)$ with this example, but not able to do so.

Answer 1

$f$ is necessarily positive in a deleted neighborhood of $0$. Also $$ |x| \ge f(x) f(2x) \ge f(x) |2x|^\alpha $$ implies $$ 0 \le f(x) \le 2^{-\alpha} |x|^{1-\alpha} $$ so that $\lim_{x \to 0} f(x) = 0$.

Answer 2

$|x|\geq f(x)f(2x) \geq f(x)|2x|^{\alpha}$. Hence, $0 <f(x)\leq 2^{-\alpha} |x|^{1-\alpha}$. Conclusion: $f(x) \to 0$ as $ x \to 0$.