Okay so this is probably the longest proof I have written. But the idea is simple, and it's pretty much copy paste in the different directions. I'm pretty confident this time around and was wondering if I have any gaps of logic or inaccuracies since we've only started practicing continuity last week.. Also about question a, I just can't seem to find such a function! Can anyone give me a direction?
b.
We will separate it into cases.
Case 1: $f\left(x\right)\ =\ L$ for every $x ∈ R$. In this case every point is an extreme point since $f(x)$ is constant.
Case 2: There exist $a ∈ R$ such that $f\left(a\right)\ >\ L$.
By definition of limit f at $∞$, we choose $ϵ\ =\ f\left(a\right)\ -\ L$ for which there exist $M_{1}>0$ such that for every $x>M_{1}$ we have $\left|f\left(x\right)-L\right|<f\left(a\right)-L$.
$\ 2L-f\left(a\right)<f\left(x\right)<f\left(a\right)$ ➜ $f\left(x\right)<f\left(a\right)$.
Note that $a≤M_{1}$ since for every $x>M_{1}$ we have $f\left(x\right)<f\left(a\right)$.
By definition of limit f at $-∞$, we choose $ϵ\ =\ f\left(a\right)\ -\ L$ for which there exist $M_{2}<0$ such that for every $x<M_{2}$ we have $\left|f\left(x\right)-L\right|<f\left(a\right)-L$.
$\ 2L-f\left(a\right)<f\left(x\right)<f\left(a\right)$ ➜ $f\left(x\right)<f\left(a\right)$.
Note that $a≥M_{2}$ since for every $x<M_{2}$ we have $f\left(x\right)<f\left(a\right)$.
Since f is continuous in R, in particular, it's continuous on $[M_{2},M_{1}]$. Then by Weierstrass's Theorem there exist $k∈[M_{2},M_{1}]$ such that $f\left(k\right)\ge f\left(x\right)$ for every $x∈[M_{2},M_{1}]$.
Since $f\left(k\right)\ge f\left(a\right)$ in $[M_{2},M_{1}]$ and $f\left(a\right)\ >\ f\left(x\right)$ for every $x$ in $(-∞,M_{2}),(M_{1},∞)$ then $f\left(k\right)\ \ge\ f\left(x\right)$ for every $x∈R$, and by definition an extreme point of $f$.
Case 3: There exist $b ∈ R$ such that $f\left(b\right)\ <\ L$.
By definition of limit f at $∞$, we choose $ϵ\ =L\ -\ f\left(b\right)$ for which there exist $M_{1}>0$ such that for every $x>M_{1}$ we have $\left|f\left(x\right)-L\right|<L\ -\ f\left(b\right)$ ➜ $f\left(b\right)\ <\ f\left(x\right)$.
Note that $b≤M_{1}$ since for every $x>M_{1}$ we have $f\left(x\right)>f\left(b\right)$.
By definition of limit f at $-∞$, we choose $ϵ\ =L\ -\ f\left(b\right)$ for which there exist $M_{2}<0$ such that for every $x<M_{2}$ we have $\left|f\left(x\right)-L\right|<L\ -\ f\left(b\right)$ ➜ $f\left(b\right)\ <\ f\left(x\right)$.
Note that $b≥M_{2}$ since for every $x<M_{2}$ we have $f\left(x\right)>f\left(b\right)$.
Since f is continuous in R, in particular, it's continuous on $[M_{2},M_{1}]$. Then by Weierstrass's Theorem there exist $s∈[M_{2},M_{1}]$ such that $f\left(s\right)\le f\left(x\right)$ for every $x∈[M_{2},M_{1}]$.
Since $f\left(s\right)\le f\left(b\right)$ in $[M_{2},M_{1}]$ and $f\left(b\right)\ <\ f\left(x\right)$ for every $x$ in $(-∞,M_{2}),(M_{1},∞)$ then $f\left(s\right)\ \le\ f\left(x\right)$ for every $x∈R$, and by definition an extreme point of $f$.
In conclusion, we have proven f must have an extream point.
Instead of working with $g(x):= f(x)-L$, we can suppose WLOG that $L=0$. If $f=0$, the proof is straightforward. If $f\neq 0$, there is $a\in \mathbb R$ s.t. $f(a)\neq 0$. Suppose WLOG that $f(a)>0$. Since $f(x)\to 0$ when $x\to \pm \infty $, there is $N>0$ s.t. $0\leq f(x)<f(a)$ for all $x\notin [-N,N]$. Since $f$ is continuous, it reach his maximum on $[-N,N]$, which is a maximum on $\mathbb R$.