‎If ‎$‎f‎$ ‎is ‎convex ‎and ‎$‎\delta‎>0‎$‎, then ‎$‎f(x + ‎\delta‎) - f(x)‎$ ‎is ‎increasing.‎

Question

My question is how to prove the following assertion:

‎‎If $f : I‎‎\rightarrow\mathbb{R}$ be a convex function, ‎$‎I‎$ ‎is ‎unbounded ‎above ‎and ‎‎$‎‎\delta‎>0‎$‎, then ‎$‎f_{‎\delta‎} : I‎‎\rightarrow\mathbb{R}$ defined by the equation ‎‎‎‎‎$‎‎‎f_{‎\delta‎} = f(x + ‎\delta‎) - f(x)‎‎‎$‎‎‎‎‎‎ ‎for ‎‎$‎x\in ‎I‎$, ‎is ‎increasing.‎‎‏ ‎‎‎Please ‎guide ‎me?‎ ‎

I know that ‎$‎f:(a, b)‎‎\rightarrow ‎‎\mathbb{R}‎‎‎‎$ ‎is ‎convex ‎if ‎and ‎only ‎if ‎there ‎is ‎an ‎increasing ‎function‎ $‎g:(a, b)‎‎\rightarrow ‎‎\mathbb{R}‎$ ‎and a‎ ‎point ‎‎$‎c\in (a, b)‎$ ‎such ‎that ‎for ‎all ‎$‎x\in (a, b)‎$‎, ‎‎$‎f(x) - f(c) = \int_{c}^x g(t) dt‎$‎. ‎

Answer 1

Let $x_1<x_2$ we want to prove that $$f(x_1+\delta)-f(x_1)\leq f(x_2+\delta)-f(x_2)$$ i.e. $$f(x_2)-f(x_1)\leq f(x_2+\delta)-f(x_1+\delta)$$ i.e. $$\frac{f(x_2)-f(x_1)}{x_2-x_1}\leq\frac{f(x_2+\delta)-f(x_1+\delta)}{x_2-x_1}\quad(*)$$ This inequality will follow by applying the increasing-chord-theorem for convex functions twice.

Theorem. Let $f$ be a convex function in some interval $I$, and let $a<b<c$ be points in $I$. Then $$\frac{f(c)-f(a)}{c-a}\leq\frac{f(b)-f(a)}{b-a}\leq\frac{f(c)-f(b)}{c-b}$$

In our case, if $x_2<x_1+\delta$, then the theorem implies that $$\frac{f(x_2)-f(x_1)}{x_2-x_1}\leq\frac{f(x_1+\delta)-f(x_2)}{x_2-\delta}$$ but since $x_1+\delta<x_2+\delta$ we can continue this inequality by applying the theorem once again, to obtain $$\frac{f(x_1+\delta)-f(x_2)}{x_2-\delta}\leq \frac{f(x_2+\delta)-f(x_1+\delta)}{(x_2+\delta)-(x_1+\delta)}$$ and so combining these two inequalities we get $(*)$. If, on the other hand, $x_1+\delta\leq x_2$, then a similar argument, with a slightly different ordering of the points, proves the inequality $(*)$ again.

Answer 2

Here is an outline of a proof:

Let $g(x)= f(x+\delta) - f(x)$.

Suppose $g$ is not increasing. There must exist $x_1$ and $x_2$ such that $x_1<x_2$ and $g(x_1)\geq g(x_2)$.

By strict convexity, $(x_2+\delta, f(x_2+\delta))$ is above the line through $(x_1, f(x_1))$ and $(x_1+\delta, f(x_1+\delta))$. (Prove that. Drawing pictures will help.)

Now, I think you can prove that $(x_2, f(x_2))$ is above the line connecting $(x_1, f(x_1))$ and $(x_1+\delta, f(x_2+\delta))$.

But that violates convexity.

That contradiction implies that $g$ is increasing.

You can try to fill in the details and let us know how it goes.

Answer 3

Let $h(x)=f(x+\delta)-f(x)$. Then for $x<x'$, $$\begin{align}h(x')-h(x)&=f(x'+\delta)-f(c)+f(c)-f(x')+f(x)-f(c)+f(c)-f(x+\delta) \\&=\int_{c}^{x'+\delta}g(t)\mathrm dt-\int_{c}^{x'}g(t)\mathrm dt+\int_{c}^{x+\delta}g(t)\mathrm dt-\int_{c}^{x}g(t)\mathrm dt \\ &=\int{x'}^{x'+\delta}g(t)\,\mathrm dt-\int{x}^{x+\delta}g(t)\,\mathrm dt\\&= \int_0^\delta\underbrace{(g(x'+t)-g(x+t))}_{\ge0}\,\mathrm dt\\&\ge 0\end{align}$$

Answer 4

A useful characterisation of convex function is given by https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable which states that $f$ is convex iff $x \mapsto R(x,y)$ is non decreasing. (Also, $R$ is symmetric.)

Now choose $x_1 \le x_2$, then $R(x_1+\delta,x_1) \le R(x_1+\delta,x_2) \le R(x_2+\delta,x_2) $, or equivalently $f_\delta(x_1) \le f_\delta(x_2)$.