$g : [0,1]\to\Bbb R$ is a concave function with $g(0) =0$ and $g(1)= \beta$. Show that $g(z) \geq \beta z$, $z \in [0,1]$.

Question

I came across the following, let $g : [0,1] \to \Bbb R$ be a concave function with $g(0) =0$ and $g(1)= \beta$. It implies $g(z) \geq \beta z$, $z \in [0,1]$. Why is the statement $g(z) \geq \beta z$ true?

Answer 1

The secant line from $[0,1]$ is

$$y-0= {\beta -0\over 1-0}(x-0)$$

By definition of concavity, the secant line is below the graph of the function on any interval, so the result immediately follows.

Answer 2

By definition of concavity on interval $[0,1]$, for any $\forall \lambda \in [0,1]\,$:

$$ g\big((1 - \lambda) \cdot 0 + \lambda \cdot 1\big) \;\ge\; (1-\lambda) \cdot g(0) + \lambda\cdot g(1) $$

Writing the above for $\lambda = z \in [0,1]\,$ with $\,g(0)=0\,$ and $\,g(1)=\beta\,$ gives $\,g(z) \ge z \cdot \beta\,$.