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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 $$F(\lambda x+(1-\lambda)y)\geq \lambda\,F(x) + (1-\lambda)\,F(y)$$

erin
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2 Answers2

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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 $[0,1)$.

ir7
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  • @erin You are welcome. If you're comfortable with it, you may want to check it (upper left corner) as accepted. Accepting an answer is relevant for this website performance statistics. – ir7 Oct 30 '14 at 12:42
  • This is really stupid but how did you get g′(z)=[zxf′(zx)−f(zx)]/z^2 – erin Nov 01 '14 at 23:52
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    Basically, from $g'(z)=\left(\frac{h(z)}{k(z)}\right)'=\frac{h'(z)k(z)-h(z)k'(z)}{k^2(z)}$, where $h(z)=f(zx)$ and $k(z)=z$ for all $z$. And remember that $x$ is a constant here. – ir7 Nov 02 '14 at 02:00
  • Thanks. That was really silly of me. my brain has turned to mush – erin Nov 02 '14 at 04:27
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Hint for the case $x=1$: $\displaystyle g(z) = \frac{f(z)-f(0)}{z-0}$. What is that the slope of?