Proving a function is quasi-concave but not concave.

Question

Can someone please help me complete my proof for the following? I am a bit confused. Thank you so much.

$\def\R{{\mathbb R}}$

We have some real-valued function $f\colon A \to \R$ where $A\subseteq \R^n$ is a convex subset, is concave if, for any $x,x'\in A$ and $\alpha \in [0,1]$, $$f((1-\alpha)x + \alpha x') \ge (1-\alpha)f(x) + \alpha f(x').$$ Moreover, $f$ is quasi-concave if, for any $x, x' \in A$ and $\alpha \in [0,1]$, $$f((1-\alpha)x + \alpha x')\ge \min\{f(x),f(x')\}.$$

i. Prove directly that $f(x) = x^2$, where $x\in \R_+$, is quasi-concave but not concave.

Answer 1

If $ 0 \leq x \leq x'$, then for all $\alpha \in [0,1]$, one has $(1-\alpha)x + \alpha x' \geq x$. Because $f : x \mapsto x^2$ is increasing on $\mathbb{R}_+$, then $$f((1-\alpha)x + \alpha x') \geq f(x) = \min \lbrace f(x), f(x') \rbrace$$

The function, however, is not concave, but convex. Indeed for all $x \in \mathbb{R}$, $f''(x) \geq 0$.

Answer 2

Pick $x=0$ and $x'=1$. Then we have for $\alpha\in (0,1)$ $$ f(\alpha) = \alpha^2 < \alpha = (1-\alpha) \cdot 0 + \alpha \cdot 1 = (1-\alpha) f(0) + \alpha \cdot f(1). $$ Hence, $f$ is not concave on $\mathbb{R}_+$. Quasi-convexity follows from monotonicity of $f$.

Added: We can do it also in the inside. If we pick $\alpha=1/2$ and $x,x'>0$ we get $$ f((1-\alpha)x + \alpha x') = f((x+x')/2)= \frac{x^2+(x')^2 + xx'}{4} \leq \frac{x^2+ (x')^2+ x^2/2 + (x')^2/2}{4} < \frac{2x^2+ 2(x')^2}{4} = \frac{x^2}{2}+ \frac{(x')^2}{2} = (1-\alpha) f(x) + \alpha f(x'). $$