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Is $$\log \log n \times \log \log \log n = \Omega(\log n) $$ How can we prove it.

Actually I'm trying to prove that $f(n) = \lceil(\log \log n)\rceil !$ is polynomially bounded. It means

$$c_1 n^{k_1} \leq f(n) \leq c_2 n^{k_2} \quad \forall n > n_0$$ $$m_1 \log n \leq \log [f(n)] \leq m_2 \log n$$ $$\log [f(n)]=\theta(\log n) \text{ i.e. } \log [f(n)]=\Omega(\log n) \text{ and }\log [f(n)]=O(\log n) $$

I've proved that $\log [f(n)] = O(\log n)$, But I'm having trouble proving $\,\log \left[f(n)\right] = \Omega\left(\log n\right)$. Can anybody tell me how can we do it.

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    $\begingroup$ The claim is the title is wrong: $\log\log n \cdot \log\log\log n = o(\log n)$. $\endgroup$ – Clement C. Aug 15 '15 at 12:22
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    $\begingroup$ Can you define the symbols you use? $\endgroup$ – Paolo Leonetti Aug 15 '15 at 12:22
  • $\begingroup$ @ClementC. f(n) = o(g(n)) then f(n)= O(g(n)) also. $\endgroup$ – Atinesh Aug 15 '15 at 12:25
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    $\begingroup$ @Atinesh Yes. But $\omega(f(n))$ does not mean what you seem to think it means. $\endgroup$ – Clement C. Aug 15 '15 at 12:29
  • $\begingroup$ (I also suspect you wanted $\Omega(\cdot)$ instead of $\omega(\cdot)$; the claim is still false then, however.) $\endgroup$ – Clement C. Aug 15 '15 at 12:38
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You will not be able to prove this. $f(n)$ grows asymptotically slower than any polynomial, i.e. $f(n) = n^{o(1)}$. Indeed, for any $c > 0$ $$ n^c = 2^{c\log n} $$ while $$f(n) = 2^{\Theta(\log\log n\cdot \log\log\log n)}$$ as you showed. But $$ \frac{\log\log n\cdot \log\log\log n}{\log n} < \frac{(\log \log n)^2}{\log n} \xrightarrow[n\to\infty]{} 0 $$

Edit: this does not contradict the fact that $f(n)$ is polynomially bounded. It is: it is asymptotically bounded (above) by any polynomial.

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  • $\begingroup$ Then How can we prove that ⌈(log logn)⌉! is polynomially bounded. $\endgroup$ – Atinesh Aug 15 '15 at 12:40
  • $\begingroup$ You did already: showed a polynomial upper bound upper bound (you wrote you did prove $\log f(n) = O(\log n))$ already). The lower bound, however, does not hold (but you do not need it). $\endgroup$ – Clement C. Aug 15 '15 at 12:41
  • $\begingroup$ I'm really Sorry, there has been a typing mistake I didn't recognize it. Please see the updated question. $\endgroup$ – Atinesh Aug 15 '15 at 14:52
  • $\begingroup$ The same proof applies -- if $\log f(n) = o(\log n)$ (and it is), then you have $\log f(n) = O(\log n)$ but cannot have $\log f(n) = \Omega(\log n)$. $\endgroup$ – Clement C. Aug 15 '15 at 14:57
  • $\begingroup$ I think I'm having a little bit confusion. Does Polynomially Bounded means Upper bounded by polynomial or lower and upper bounded by polynomial. $\endgroup$ – Atinesh Aug 15 '15 at 15:02

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