I want to know if the below two function are Linear or not. I have been searching in Google but found some confusing results. The first one is an inequality and the second one is just a function and both are function of $w$.
$[1.] -(w-1) \frac{Z g p_{max} + \sigma^2}{g_{pu} p_{pu}} \leq \log q$
$[2.] f(w) = -\alpha' \log_2(w)$
Any explanation will be highly appreciated. thanks.
On
I will only give an answer regarding the second item.
Even if most people will contend that $f(w) = -\alpha' \log_2(w)$ was clearly non-linear, this is not so if you consider the domain and the image of $f$ to be vector spaces. Generally, ...
... a function $f : V \to W$ between two vector spaces $(V,\oplus,\odot)$ and $(W,\boxplus,\boxdot)$ over a field $F$ (the set of scalars) is linear, if and only if it preserves the vector addition and the scalar multiplication.
So, you would have to show that $$ f(a \oplus b) = f(a) \boxplus f(b) \quad \text{and} \quad f(\lambda \odot a) = \lambda \boxdot f(a) $$ for all $a,b \in V$ and $\lambda \in F$.
Claim: Let $(V,\oplus,\odot)$ and $(W,\boxplus,\boxdot)$ be two vector spaces over a field $F$ where
- $V := \mathbb{R}_{>0}$,
- $a \oplus b := a \cdot b$,
- $\lambda \odot a := a^\lambda$,
and
- $W := \mathbb{R}$,
- $a \boxplus b := a + b$,
- $\lambda \boxdot a := \lambda \cdot a$.
Then $f : V \to W, \, w \mapsto -\alpha'\log_2(w)$ is linear.
Too see this, check the two conditions of linearity. $$ f(a \oplus b) = f(a \cdot b) = -\alpha' \log_2(a \cdot b) = -\alpha' \log_2(a) -\alpha' \log_2(b) = f(a) + f(b) = f(a) \boxplus f(b) $$ and $$ f(\lambda \odot a) = f(a^\lambda) = -\alpha' \log_2(a^\lambda) = -\alpha' \lambda \log_2(a) = \lambda \cdot f(a) = \lambda \boxdot f(a). $$ So, $f$ is linear.
The second one is clearly nonlinear. Why? Well, differentiate with respect to $w$: if it were linear, the result would be a multiple of $w$, but this is not the case.
The left-hand side of the first "thing" is linear in $w$ (it is a multiple of $w-1)$), but I am not sure if you are asking this.