Condition that maybe implies convexity

Question

We have $f:R\to R, f(x+a)-f(x) \leq f(y+a)-f(y), \forall x,y,a \in R, x\leq y, a\ge 0$

We have to prove that the function satisfies the following:

$f((1-\frac{1}{2^k})x +\frac{1}{2^k}y) \leq (1-\frac{1}{2^k})f(x)+\frac{1}{2^k}f(y), \forall x,y \in R, \forall k$ positive integer. That inequality is reminiscent of the inequality for convex functions. Is there a connection? How could I approach the problem?

Answer 1

For given $x < z$ you can set $y = \frac{x+z}{2}$ and $a=\frac{z-x}{2}$ in $$ f(x+a)-f(x) \leq f(y+a)-f(y)$$ to get $$ 2 f(\frac{x+z}2) \le f(x) + f(z) \, , $$ i.e. $f$ is midpoint-convex. That is the desired inequality for $k=1$, and the general case follows by induction, since $$ (1-\frac{1}{2^{k+1}})x +\frac{1}{2^{k+1}}y = \frac 12 x + \frac 12 \left((1-\frac{1}{2^k})x +\frac{1}{2^k}y\right) $$

(For continuous functions, midpoint-convexity implies convexity.)

Answer 2

Let $y\geq x$ and choose $a=y-x \geq 0$, than $$ f(y)-f(x) \leq f(2y)-f(y) \quad \iff \quad 2 f(y) \leq f(2y-x)+f(x). $$

Now I prove the statement by induction on $k$.

• $k=1$, the inequality above gives $$ f(y) \leq \frac{1}{2} f(2y-x) + \frac{1}{2} f(x). $$ Choose $y = \frac{1}{2} a + \frac{1}{2}b$, we get $$ f\left( \frac{1}{2} a + \frac{1}{2}b \right) \leq \frac{1}{2} f(a+b-x) + \frac{1}{2} f(x). $$ Choosing $x = \min\{a,b\}\leq \frac{1}{2}a + \frac{1}{2}b$, we get the sought inequality $$ f\left( \frac{1}{2} a + \frac{1}{2}b \right) \leq \frac{1}{2} f(a) + \frac{1}{2} f(b). $$ Assume, w.l.g, that $x=a$.
• $k \to k+1$, assuming it holds true for $k$, we know that $$ f\left( \left( 1-\frac{1}{2^k} \right)a + \frac{1}{2^k}b \right) \leq \left( 1-\frac{1}{2^k} \right) f(a) + \frac{1}{2^k} f(b). $$ Now set $y=\left( 1-\frac{1}{2^{k+1}} \right)a + \frac{1}{2^{k+1}}b $ and $x=a$ in the first inequality and use the inductive step (IS) \begin{align} f\left( \left( 1-\frac{1}{2^{k+1}} \right)a + \frac{1}{2^{k+1}}b \right) & \leq \frac{1}{2} f\left( 2a - \frac{1}{2^k}a + \frac{1}{2^k}b - a \right) + \frac{1}{2} f(a) \\ & = \frac{1}{2} f\left( a\left(1 - \frac{1}{2^k}\right) + \frac{1}{2^k}b \right) + \frac{1}{2} f(a) \\ & \overset{\text{IS}}{\leq} \frac{1}{2} \left( \left( 1-\frac{1}{2^k} \right) f(a) + \frac{1}{2^k} f(b) \right)+ \frac{1}{2} f(a) \\ & = \left( 1-\frac{1}{2^{k+1}} \right) f(a) + \frac{1}{2^{k+1}} f(b) \end{align}