# Squeeze-defining a function

If for all real numbers $3>a>5/2>b>2$, there exist $e>0$ such that $x^b<f(x)<x^a$ for all $x>e$, must $f(x)=x^{5/2}$?

Followups:
If for all real numbers $3>a>5/2>b>2$, $x^b<f(x)<x^a$ for all $x$, must $f(x)=x^{5/2}$?

If for all real numbers $e>0$ and $3>a>5/2>b>2$, $x^{b-e}<f(x)<x^{a+e}$ for all $x$, must $f(x)=x^{5/2}$?

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You probably wish to require that $f$ is continuous too. – Asaf Karagila Oct 15 '11 at 23:04
I like that you're following up your question with related questions, but I think it would be a good idea to inform the answerers about the follow-ups. People sometimes 'move on' after answering the question (unless they are improving the answer), and there's some chance I did not notice that you added follow-ups. – Srivatsan Oct 15 '11 at 23:23

Original question. No. The function $x^{5/2} (\ln x)^{k}$ (for any real $k$) is a counterexample.
Follow-up question 1. The answer is yes. Suppose $x^b < f(x) < x^a$ for all $b < 5/2 < a$. Note that this hypothesis cannot hold for any $x \leq 1$, so we'll assume $x > 1$.
Fix any $x > 1$. Then we have: $$b \ln x < \ln f(x) < a \ln x \ \ \implies \ \ b < \frac{\ln f(x)}{\ln x} < a,$$ for all $b < 5/2 < a$ (remember that $\ln x > 0$). Since $\frac{\ln f(x)}{\ln x} < a$ for all $a > 5/2$, it follows that $\frac{\ln f(x)}{\ln x} \leq 5/2$ (why?). We can similarly show $\frac{\ln f(x)}{\ln x} \geq 5/2$. Hence $$\frac{\ln f(x)}{\ln x} = \frac 52 \ \ \iff f(x) = x^{5/2} ,$$ which is what we wanted.
It is an interesting exercise to understand why the above argument fails in the case of the counterexample $x^{5/2} (\ln x)^k$.
Followup question 2. If $x^{b-e} < f(x) < x^{a+e}$ for all $b < 5/2 < a$ and all $e > 0$, then we can show that the seemingly stronger condition $x^{b} < f(x) < x^{a}$ also holds for all $b < 5/2 < a$. (Exercise!) So this question is the same as the previous follow-up question.