Background (Quasiconvex defn):
A function $f: S \rightarrow \mathbb{R}$ defined on a convex subset $S$ of a real vector space is quasiconvex if for all $x, y \in S$ and $\lambda \in[0,1]$ we have $$ f(\lambda x+(1-\lambda) y) \leq \max \{f(x), f(y)\}$$
Now I'm interested in showing this for
$ |x|^{p}$ for $0<p<1$ , $S = \mathbb{R}$
It is essentially Wikipedia's image of a quasiconvex function
I can see how it is true by looking at the graph, If I draw any line from two points on the graph, one of the ends of the line will be greater than the rest of the line.
Now I don't know how to proove this, essentially prooving$$ |\lambda x+(1-\lambda) y|^{p} \leq \max \{|x|^{p}, |y|^{p}\}$$.
I'm not sure if this is rigorous enough or correct.
But using the hint, I think it prooves it for any function that satisfies the hint.
For any $f: \mathbb{R} \rightarrow \mathbb{R}$, we know $\mathbb{R}$ is convex set. If f is monotonically increasing in $[0,\inf)$ and decreasing in (-inf,0].
Then for any two points x,y
Either x,y are both positive or both negative than monotonicity is just the same as below but one side.
Or one is positive one is negative.
W.L.O.G Then if x $\in (-\inf,0]$ and y $\in [0,\inf) $then for$ z \in (x,y) $
$z \in (x,0] $
f(x) > ... f(z) .. ≥f(0)
$z\in [0,y)$
f(0) ≤.. f(z) <... f(y).
Therefore max f(z) st. z $\in [x,y]$ is either f(x) or f(y), and is greater than any f(z) = (+(1−)) $0<\lambda<1$.
Proof that f satisifies the hint in the case of interest, we can use the derivative: $$df/dx = p*x*|x|^{p-2}$$ 0<p<1, negative for negative x, positive for positve x