It's my first post here so be gentle ^^
My problem formulates like this:
Given the functions $f,g:\mathbb {N} \to \mathbb {R_0^+}$ with the property $g(n) \not = 0 \ \ \ \forall n\in \mathbb {N}$, is the following equivalence true ?
$∃α > 0 : ∀n ∈ N : f(n) ≤ α · g(n)$
$\Leftarrow\Rightarrow$
$f \in\{f \ | \ ∃\ α > 0 , ∃ \ n_0 ∈ \mathbb{N} \ , \ ∀n ≥ n_0 : 0 ≤ f (n) ≤ αg(n)\}$ (This set will be reffered to as $O(g)$)
Since this is all quite theoretical I tried to make a case and get a feel for the problem.
So I asked my self a new question:
Given that $1/x \leqslant x+c$ with $c$ being some constant, holds for all $x\geqslant 1+c$ we know $\\$ $f(x)=1/x$ and $g(x)=x+c$ suffices $f \in O(g) $ with $n_0 = 1$, we should be able to pinpoint the biggest $f(x)$ for $x<1$ as the superior of the following set: $I:=\{f(x) \ : \ f <g \text{ with } x >1 \ \land f <g \text{ with } x <1 \}$ Let`s call $sup(I)$ = $\epsilon$
With the help of $\epsilon$ it should be possible to somehow algeabraically find an $\alpha$ so that $\alpha g(x) = \epsilon$ How the algebra works in that case is not interesting to my problem but if that holds $f$ and $g$ suffice the first part of the equivalence on the top. And that would be the first step to get a proof for general $f$ and $g$ sufficing said equivalence.
Thank you for reading and I`m looking forward to creating a nice proof with you guys
For the stray soul who will be needing an answer in the future like I did in the past:
Let $f,g :\mathbb{N \to R_0^+} \text{ and } g(n) \not = 0 ,∀n \in \mathbb{N}$. Then : $f \in O(g)<==> ∃\ α > 0 : ∀n \in \mathbb{N}: f(n)≤α*g(n) $
Proof:$ \ $ "==>"
Let $ f \in O(g)$ <==>$ ∃ n_0 \in \mathbb{N} , ∃ α > 0 :0≤f(n)≤ g(n)*α ,\forall n≥ n_0 \ $
Let $N := \{n < n_0\}$, $N$ contains $n_0$ elements and is therefore finite. Let $m_f = max ( f(N))$ and $m_g = min (g(N))$
Then : $\frac{1}{m_f} *f(N)≤1 $ and $m_g * g(n)≤1 $
Therefore is $\forall n \in \mathbb{N}$:
$\frac{1}{m_f}*f(n)≤m_g *g(n) <==> f(n)≤m_f *m_g*g(n) $
Choosing $\beta = max(\alpha , m_g*m_f$
Then: $0 ≤ f(n) ≤ \beta *g(n), \forall n \in \mathbb{N}$
"<==":
Let $∃\ α > 0 : ∀n \in \mathbb{N}$ be true:
Going after the premise: $f(n) ≥ 0, \forall n \in \mathbb{N}$
Then for $n_0=0$: $\exists \alpha > 0 : 0 ≤ f(n) ≤ \alpha * g(n)$
$End$ $ of$ $ Proof$