Let $f : \mathbb{R}_+ \to \mathbb{R}$ be a (differentiable) function that satisfies the following
- $f(x) = 0\iff x =0$
- $\forall x > 0, f(x) \geq 0$
- $f$ is stricly concave
Show that $f$ is monotonically increasing
My lecturer did the following in proof class
Lecturer's Proof: By concavity of $f$ we have $f(x) < f'(x_0)(x-x_0) + f'(x_0)$ for all $x \in [0, \infty)$. Suppose $f$ was not monotonically increasing, then there exists an $x_0 \in [0, \infty)$ for which $f'(x_0) < 0$, hence there exists an $x \in \mathbb{R_+}$ such that $f(x) < 0$ (using concavity of $f$) contradicting positivity of $f$.
But I am having trouble understanding one part of the proof. How does once conclude by concavity of $f$ we have $f(x) < f'(x_0)(x-x_0) + f'(x_0)$ for all $x \in [0, \infty) = \mathbb{R_+}$?
Note that for a concave function the tangent line at any point on the graph of the function doesn't lie below the graph. But the equation of a tangent line at any point $x_0\in\text{dom} f$ (domain of $f$) is given by $$f'(x_0)(x-x_0)+f(x_0)$$ i.e. the first order Taylor approximation of $f$ at $x=x_0$. Since function is concave by the above geometric property of its tangent lines it must be the case that for any $x\in\text{dom}f$ we must have $$f(x)\leqslant f'(x_0)(x-x_0)+f(x_0)$$