$\,f:[0,1)\to\mathbb R$ is a concave differentiable function st $\,f(0)=0$. Show that $g:[0,1),g(z)=f(zx)/z$, for $x > 0$ is decreasing Question:
Suppose that $f : [0, 1) \to\mathbb R$ is a concave differentiable function such that $\,f(0) = 0$. Show that $g : [0, 1) \to\mathbb R$ defined by
$g(z) = f(zx)/z$, for some given $x > 0$, is a decreasing function of $z$.
Thoughts: I am having quite a lot of trouble with this question and am unsure where to start. A concave function satisfies
$$F(\lambda x+(1-\lambda)y)\geq \lambda\,F(x) + (1-\lambda)\,F(y)$$
 A: Hint:
As $f$ is concave and differentiable, it should lie below all its tangents, that is:
$$ f(b)\leq f(a)+f'(a)(b-a) $$
for all $a,b \in[0,1)$. 
On the other hand, for $g$ (easily seen to be differentiable on $(0,1)$) we have:
$$g'(z) = \frac{zxf'(zx)-f(zx)}{z^2} $$
which can be proven (using above observation and $f(0)=0$) to be less than or equal to $0$ for all $z\in (0,1)$, hence $g$ is decreasing on $(0,1)$. 
Note: 
Note that $g(0)$ doesn't exist, so one might want to restrict the domain of $g$ to $(0,1)$. Alternatively, one can define $$g(0) := \lim\limits_{t\to 0}g(t) = \lim\limits_{t\to 0}\frac{f(tx)-f(0)}{t-0}=xf'(0),$$
in which case:
$$ g(0)= xf'(0) \geq g(z)=\frac{f(zx)}{z}$$
for all $z\in (0,1)$, again due to the initial observation. So, in this case, $g$ is decreasing on $[0,1)$.
Note also that one should have $x\in (0,1]$ or should enlarge the domain of $f$, say to $[0,\infty)$, otherwise one can't apply $f$ to $zx$ as it may not be in $[0,1)$. 
A: Hint for the case $x=1$: $\displaystyle g(z) = \frac{f(z)-f(0)}{z-0}$.  What is that the slope of?
