$f''(x)\leq 0$ implies $f$ concave

I'm using the following definition:

$$f:[0,\infty)\rightarrow[0,\infty)$$ is concave if $$\forall x,y\in[0,\infty)$$ and $$s\in[0,1]$$, we have

$$f(sx+(1-s)y)\geq sf(x)+(1-s)f(y)$$

I need to prove that every function $$f:[0,\infty)\rightarrow[0,\infty)$$ twice differentiable satisfying $$f''(x)\leq 0$$ for all $$x\in[0,\infty)$$ is concave.

I found the reciprocal, but not this statement.

Well, I know that $$f''(x)\leq 0$$ implies that for every $$x, $$f'(y)\leq f'(x)$$. Can someone give me just some hints?

• See this. The answers use techniques you should use as well. Dec 4 '18 at 17:49

Use Rolle's Theorem.

Define $$z = sx + (1-s)y$$.

$$\exists z_1 \in (x, z), \quad\text{s.t.}\quad f(z) - f(x) = (z-x)f'(z_1)$$

$$\exists z_2 \in (z, y), \quad\text{s.t.}\quad f(y) - f(z) = (y-z)f'(z_2)$$

so \begin{aligned} sf(x) + (1-s)f(y) = & s[f(z) - (z-x)f'(z_1)] + (1-s)[f(z) + (y-z)f'(z_2)] \\ = & s[f(z) - (1-s)(y-x)f'(z_1)] + (1-s)[f(z) + s(y-x)f'(z_2)] \\ = & [s+(1-s)]f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)]\\ = & f(z) + s(1-s)(y-x)[f'(z_2) - f'(z_1)] \end{aligned}

Apparently $$s>0, 1-s>0, y-x>0$$ and also $$f'(z_2) \leq f'(z_1)$$ because $$f''(x)$$ always negative.

So $$[sf(x) + (1-s)f(y)] - f(z) = s(1-s)(y-x)[f'(z_2)-f'(z_1)] \leq 0$$