A basic question about concave function.

Question

Question:

Let $f: \mathbb{R}_{+}^{n} \rightarrow \mathbb{R}$ be a concave function satisfying $f(0)=0 $. Show that for all $k \geq 1 $ we have $k f(x) \geq f(k x) $. What happens if $ k \in[0,1) ?$

I know that when $ k \in [0,1)$, by the definition of concave function and the condition given, $f(0) = 0$, we can show that $kf(x) \leq f(kx)$. But when $k \geq 1,$ I don't know how to prove this conclusion.

Answer 1

So you could deal with the $k<1$ case because $kx$ was in between $0$ and $x$ and you could apply concavity.

Now if $k\geq 1$ which of $0,kx$ and $x$ is between the two others ? What can you deduce ?