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If $f(n)=\Omega(g(n))$ and $h(n)=\theta (g(n))$ then does this implies $f(n).g(n)=\Omega(g(n).h(n))$

I saw a proof where they have proved If $f(n)=O(g(n))$ and $h(n)=\theta (g(n))$ then this implies $f(n).g(n)=O(g(n).h(n))$.

So I thought may this should also hold. I have taken some examples and the formula holds on all of them. However I can't be sure. There can be a contradicting case.

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  • $\begingroup$ Use the definitions. $\endgroup$ – Raphael Dec 31 '17 at 9:34
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I have got the answer. It's easy.

Since $f(n)= \Omega g(n)$

$cg(n)\le f(n)$

Multiplying both sides by h(n)

$cg(n).h(n)<=f(n).h(n)$

$f(n).g(n)=\Omega(g(n).h(n))$

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  • $\begingroup$ The penultimate inequality depends on whether $h(n)\ge 0$. In addition, the right hand side is $f(n)h(n)$ while the left hand side of the last equality is $f(n)g(n)$. $\endgroup$ – xskxzr Mar 1 '18 at 14:57

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