I'm trying to prove the following lemma:$c$ is a positive real number and $f, g$ are functions from natural numbers to non-negative real numbers. I'm trying to prove rigorously that:$\Omega(cf(n))$ = $\Omega(f(n))$.I know that it is obvious but I'm trying to construct a proof that is as complete as possible. My current approach is like this:This lemma is equivalent to saying that: $f(n) \in \Omega(cf(n))$ iff $cf(n) \in \Omega(f(n))$.We can also restate that as:If $t(n) \in \Omega(cf(n))$ then $t(n) \in \Omega(f(n))$.If $t(n) \in \Omega(f(n))$ then $t(n) \in \Omega(cf(n))$.For 1.,$(*)$ $\exists d_1, d_2 \gt 0, \forall n \gt n_0, n_1, \forall n \in N$:$t(n) \ge d_1cf(n)$ and $t(n) \ge d_2f(n)$Now let's fix $d_1, d_2$ and $n_0, n_1$ to be any constants that fulfils $(*)$, such that:$n'=max\{n_0, n_1\}$ and $d_1c \ge d_2$, using this we can say that:$t(n) \ge d_1cf(n) \ge d_2f(n)$ and hence the 1. is satisfied because $t(n) \in \Omega(cf(n)), \Omega(f(n))$. The proof of 2. is mutatis mutandis.Do I have a mistake in my proof, is there a better/more elegant way to prove this lemma? Shortly how can I improve this?
Proving $\Omega(cf) = \Omega(f)$
asymptotics;proof techniques;check my proof