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Given a function $g:(0,1) \to (0,\infty)$ such that $\lim\limits_{x \to 0}{\frac{g(x)-1}{x}}=0$, I try to show that this implies $\lim\limits_{k\to\infty}{g(x2^{-k})^{2k}}=1$. Using Taylor-Expansion is no alternative, since I don't know anything about differentiability . I've tried using $ g(x)=o(x)+1$ and applying the logarithm as well as rewriting $g(x2^{-k})$ by adding ones, but it doesn't quite work out.

Does anyone know some easy trick to show this?

rhodelta
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2 Answers2

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I assume $x\in (0,1)$ is fixed. From what you wrote we can say $g(x/2^k) = 1 + o(x/2^k)$ as $k\to \infty.$ Thus given $\epsilon>0,$ we have $g(x/2^k) \le 1 + \epsilon (x/2^k)$ for large $k.$ Therefore $g(x/2^k)^{2^k} \le (1 + (\epsilon x)/2^k)^{2^k}$ for such $k.$ It follows that

$$\limsup_{k\to \infty} g(x/2^k)^{2^k} \le e^{\epsilon x}.$$

Since $\epsilon$ was arbitrary, the above $\limsup$ is $\le 1.$ There is a similar argument from below.

zhw.
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  • Thanks for your help, it is much appreciated. I guess in order to show the statement I still need to show $\liminf\limits_{k\to\infty} g(x/2^k)^{2^k} \ge 1$. I am thinking this can be shown by using $g(x2^{-k}) \ge - g(x2^{-k}) \ge -1-\epsilon x 2^{-k}$ (since $g$ maps to $(0,\infty)$) and thus $g(x2^{-k})^{2^k} \ge (-1-\epsilon x 2^{-k})^{2^k} = (-1)^{2^k} (1+ \epsilon x 2^{-k})^{2^k}$ and then using the same arguments as above? – rhodelta Jun 09 '17 at 19:15
  • No, you have $g(x2^{-k}) = 1 +o(x2^{-k}),$ giving $g(x2^{-k}) \ge 1-\epsilon x2^{-k}$ for large $k.$ – zhw. Jun 09 '17 at 19:19
  • Well, I guess that's that. Thanks a lot! – rhodelta Jun 09 '17 at 19:43
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Use the following lemma (courtesy Thomas Andrews):

If $a_{n} $ is a sequence such that $n(a_{n} - 1)\to 0$ then $a_{n}^{n} \to 1$.

Let $a_{n} =g(x/2^{n})$ and then we can see that $$n(a_{n} - 1)=n\frac{x}{2^{n}}\cdot\frac{g(x/2^{n})-1}{x/2^{n}}\to 0\cdot 0=0$$ and hence $a_{n} ^{2n}=(a_{n}^{n})^{2}\to 1$.

Paramanand Singh
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  • Ahem ... This is basically my proof, the difference being: I proved the lemma. – zhw. Jun 10 '17 at 17:27
  • @zhw.: well you did prove the lemma, but your answer works by using exponential function. The proof given in the link in my answer shows that the lemma does not require anything about exponential /logarithm. Just plain algebra suffices. But yes on a deeper note both the answers are equivalent. And I doubt if a substantially different answer is available. – Paramanand Singh Jun 10 '17 at 17:44