If $f(n) = \mathcal O(g(n))$ then is $f(n) = \Theta(g(n))?$
My attempt:
$\exists n_0 : \forall n\geq n_0, \exists c>0: f(n)<cg(n) $
Now for large large n, if we take (c/(some integer))*g(n), its graph lies below f(n) {cf(n) (c>1>d>0) more f(n) towards y axis and df(n) takes f(n) away from y-axis }. For sufficiently large n, we can find reasonable c >>0(defenetely not small c for lower bound). So $f(n) = \Theta(g(n))?$
Can any 1 please tell me is this proof correct?
Another condition: If $f(n) = \mathcal O(g(n))$ and if **$f(n) \neq \Omega(g(n)) $**then is $f(n) = \Theta(g(n))?$
No right? please correct me if i am wrong.
can we take f(n) to be discontinuous function with f(n) = 0 when n is even and f(n) = 1 when n is odd then g(n)=1 is only the upper limit.
Does $\Omega$ or $\omega$ (asymtotically tight lower bound or asymtotically loose lower bound) mean that can we take smaller c ($c \to 0$) for lower bound? No right? If so how large must be c?