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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 $x<y$, $f'(y)\leq f'(x)$. Can someone give me just some hints?

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    $\begingroup$ See this. The answers use techniques you should use as well. $\endgroup$ Dec 4 '18 at 17:49
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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 $[sf(x) + (1-s)f(y)] - f(z) = s(1-s)(y-x)[f'(z_2)-f'(z_1)] \leq 0$

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