My question essentially concerns how to prove if a function is negligible. As a matter of fact, though this question might seem very basic, it seems that most of the "proofs" of negligible concern comparing the the function with another negligible function and then asserting its status via laws of negligible functions.
What I am insisting on is a direct mathematical proof of proving how functions can be proven as negligible without comparison with any other function.
As stated by Dan Boneh:
Different communities define these (negligible and non-negligible) differently
For practitioners they are basically a scalar $\varepsilon$, where, for example:
On the other hand, in theory, $\varepsilon$ is considered a function $\mathbb{N} \to \mathbb{R}$.
Saying that a function is non-negligible means that the function is bigger than some polynomial infinitely often:
$\exists d: \varepsilon(\lambda) \geqslant 1/\lambda^d$
Saying that a function is negligible means that the function is smaller than all polynomials:
$\forall d, \lambda \geqslant \lambda_d: \varepsilon(\lambda) \leqslant 1/\lambda^d$
where $\lambda_d$ is some integer depending on $d$ and, in both cases, $\lambda$ is an integer.
That said we can see that $\varepsilon(\lambda) = 2^{-\lambda} = 1/2^\lambda$ is negligible, because for any constant $d$ there is a sufficient large $\lambda$ so $\varepsilon(\lambda) \leqslant 1/\lambda^d$.
We can also verify the other example from the link e-sushi provided: $\varepsilon(\lambda) = 2^{-100} = 1/2^{100}$ is non-negligible because if we set $d = 1000$ this function is clearly bigger than $1/\lambda^d$
$\square$
Note: Most of the answer is based on Dan Boneh's explanation about this subject on his coursera course Note 2: This is also a nice answer on the subject
