Find $f(x)$ verifying three constraints

Question

It's just a question to understand better convexity and function :

Let $x>\epsilon$ with $\epsilon>0$ and $f(x)$ a continuous , twice differentiable function on $(\epsilon,\infty)$ verifying the following constraints : $$\forall x>\epsilon \quad f(x)<x$$ $$\forall x>\epsilon \quad f''(x)<0$$ $$\lim_{x \to \infty }(x-f(x))=0$$

Motivation

Find a function wich behaves like $f(x)=x$ but have different properties particulary with convexity .

My try

After one hour I have found the function $f(x)=x^{x\sin\Big(\frac{1}{x}\Big)}$ but it's hard to deal with the second derivative so my question :

Have you a simpler example where we can calculate easily all the constraints ?

Thanks a lot for your help and your contribution .

Answer 1

You won't be able to find a convex function satisfying those constraints. If $f$ is a convex function and $f$ is differentiable, then its second derivative must be nonnegative i.e. $f''(x)\geq0$ on its domain.

Answer 2

This may answer some things :

Theorem 1: Given any continuously differentiable function $y(x)$ and if $\frac{dy}{dx}>0 $ for $x \in [a, b]$: Then there exists a variable $\varphi_n$, and $\varphi_n(x)$ is continuously differentiable over $\mathbb{R}$ with the following properties:

i. $|\varphi_n(x)-y(x)| \le \epsilon$ for $x \in (a, b)$

ii. $\varphi_n(x)=x$ for sufficiently large $|x|$

iii. $\lim\limits_{n\mapsto 0}\varphi_n(x)=y$ and $\lim\limits_{n\mapsto 0}\frac{d\varphi_n}{dx}=\frac{dy}{dx}>0$ for $x \in (a, b)$

iv. The mapping $\varphi:x\to R$ is one-to-one, so it is invertible

Proof:

$r(x)=-{\frac {n}{x^2-a^2}}$

$h(x)=-{\frac {n}{b^2-x^2}}$

$\psi_n(x) = \begin{cases} e^{r(x)+h(x)}, & \text{if $a <x< b$ } \\ 0, & \text{otherwise} \end{cases}$

define $\varphi_n=x(1-e^{-nx^2})+\psi_n(x) y(x)+\int_{-\infty}^x2nt(\frac{-1}{(t^2-a^2)^2}+\frac{1}{(b^2-t^2)^2})\psi_n(t)y(t)dt$

it is easily seen $\frac{d\varphi_n}{dx}=(1-e^{-nx^2})+2nx^2e^{-nx^2}+\psi_n(x)\frac{dy}{dx}$

all the terms on the right hand side are greater than zero

Taking limits and using inverse function theorem in one dimension, the theorem follows