Given $a'>a,b>0$ and $f$ concave, show that $f(a+b)-f(a)\geq f(a'+b)-f(a')$.
Can anyone give me a hint on how to show that inequality using the concave definition $f(tx+(1-t)y)\geq tf(x)+(1-t)f(y)$? Thank you!
2 Answers
A general property of concavity is for $ x_1 < x_2 < x_3$
$$\frac{f(x_2)-f(x_1)}{x_2-x_1} \geqslant \frac{f(x_3)-f(x_1)}{x_3-x_1} \geqslant \frac{f(x_3)-f(x_2)}{x_3- x_2}.$$
This can be proved from the basic $f(tx + (1-t)y) \geqslant tf(x) + (1-t)f(y)$ for $t \in [0,1]$.
With
$$t = \frac{x_3-x_2}{x_3 - x_1}$$
we have $x_2 = tx_1 + (1-t)x_3$ and
$$f(x_2) \geqslant \frac{x_3-x_2}{x_3 - x_1}f(x_1) + \left[1-\frac{x_3-x_2}{x_3 - x_1}\right]f(x_3)\\ \implies \frac{f(x_3)-f(x_1)}{x_3-x_1} \geqslant \frac{f(x_3)-f(x_2)}{x_3- x_2}.$$
The other inequality is shown in a similar way using
$$t = \frac{x_2-x_1}{x_3 - x_1}.$$
Applying the three-slope inequality, if $a < a' < a+b < a' +b$ then
$$\frac{f(a+b)-f(a)}{b} \geqslant \frac{f(a+b)-f(a')}{a+b-a'} \geqslant\frac{f(a'+b)-f(a')}{b}$$
In the other case, $a < a+b < a' < a' +b$ and we have
$$\frac{f(a+b)-f(a)}{b} \geqslant \frac{f(a')-f(a+b)}{a'- a-b} \geqslant\frac{f(a'+b)-f(a')}{b}$$
Therefore,
$$f(a+b)-f(a) \geqslant f(a'+b) - f(a').$$
Here is that more direct approach which my intuition was telling me about.
It is based on the observation that the proportion in which $a+b$ divides the interval $[a, a'+b]$ is same as the proportion in which $a'$ divides the same interval (only viewed in the opposite direction). This also means that:
$\frac{(a+b) - a}{(a'+b-a)} + \frac{(a'-a)}{(a'+b-a)} = 1$.
That's because both points are at distance $b$ from one of the interval ends.
Now we don't need to look at different cases regarding which number is bigger, $a+b$ or $a'$.
What we know for sure is that they are both internal to that interval.
We can write (using the concavity property for the interval $[a, a'+b]$):
$f(a') \geqslant \frac{a'-a}{a'+b-a} \cdot f(a'+b) + \frac{b}{a'+b-a} \cdot f(a)$
(here we use $a'$ as the internal point, anyone can see what $t$ is)
Also (again using the concavity property for the interval $[a, a'+b]$):
$f(a+b) \geqslant \frac{b}{a'+b-a} \cdot f(a'+b) + \frac{a'-a}{a'+b-a} \cdot f(a)$
(here we use $a+b$ as the internal point, anyone can see what $t$ is)
Now we just add the two inequalities together and we get what we wanted to prove.
