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Following is problem 1.1.9(b) in Problems in Mathematical Analysis II.

Show that if in a deleted neighborhood of zero the inequalities $f(x)\ge|x|^{\alpha}$, $\frac12<\alpha<1$, and $f(x)f(2x)\le|x|$ hold, then $\lim_{x\to0}f(x)=0$.

The given solution used $|x|^{\alpha}\le f(x)\le\frac{|x|}{|2x|^{\alpha}}$ and concludes that $\lim_{x\to0}f(x)=0$ (probably by the Squeeze Theorem). My question is, why do we need $\frac12<\alpha<1$? Is $0<\alpha<1$ not enough?

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For concreteness, we look, for example, at $\alpha=1/4$. But the issue described below comes up for any positive $\alpha \le 1/2$.

The same argument as the one you gave yields $$|x|^{1/4} \le f(x) \le \frac{∣x∣^{3/4}}{2^{1/4}}.$$

When $0<|x|<1$, the above inequality cannot hold, since $|x|^{1/4}>|x|^{3/4}$. So it would be strange to hypothesize that it holds in a deleted neighbourhood of $0$.

It is technically correct to say that if $0<\alpha<1$, and if the inequality holds, then $\displaystyle\lim_{x\to 0}f(x)=0$. However, the part of the result that deals with $0<\alpha\le 1/2$ is not interesting, since there is no $f$ that satisfies the inequality,

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I see.. Thanks a lot! – noob Oct 31 '11 at 11:00

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