Find a convex function $f$ which satisfies $\lim\limits_{x\to+\infty}f'(x)=a$ and $\lim\limits_{x\to+\infty}\Big[x\Big(f'(x)-a\Big)\Big]\neq0$

Question

Question 1.

Find a function which satisfies the following conditions:

$f:\mathbb{R}\to \mathbb{R} ,\quad f\ \cup$

$\lim\limits_{x\to+\infty}f'(x)=a$

$\lim\limits_{x\to+\infty}\Big[x\Big(f'(x)-a\Big)\Big]\neq0$

It relates to this problem. I'm unable to find a such function.

Thanks in advance.

Edit: One final question (Perhaps what I should have asked).

Question 2.

Find a function $g$ such that:

$g:\mathbb{R}\to \mathbb{R}, \ g \ \cup $

$\lim\limits_{x\to+\infty}\Big[g(x)-ax \Big]=\lambda\in\mathbb{R}\quad$

$\lim\limits_{x\to+\infty}\Big[x\Big(g'(x)-a\Big)\Big]\neq0$

Let's note that if $\lim\limits_{x\to+\infty}\Big[g(x)-ax \Big]=\lambda\in\mathbb{R}$ $\ $then $\lim\limits_{x\to+\infty}g'(x)=a$

Also it can be shown that $\exists\lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big) $

Second Edit ((Question 2. is impossible)

set $\lambda = k+c, c\neq 0$ then

$\lim\limits_{x\to+\infty}\Big[g(x)-ax-k \Big]=c\neq 0\quad$

So, $c=\lim\limits_{x\to+\infty}\Big[g(x)-ax-k\Big]=\lim\limits_{x\to+\infty}\Big[{x(g(x)-ax-k)\over x} \Big]\overset{{\infty}\over \infty}{=}\lim\limits_{x\to+\infty}\Big[{\Big(x(g(x)-ax-k)\Big) '\over \Big(x\Big) '} \Big]=\lim\limits_{x\to+\infty}\Big[g(x)-ax-k+x(g'(x)-a) \Big]=c + \lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big)\\ \Longrightarrow \lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big)=0$

Answer 1

I'm not the most au fait with convex analysis but here is a function which I believe satisfies your conditions.

Consider the continuously differentiable function

$$f(x) = \begin{cases} x-\log(x+1) & x >0 \\ 0 &x\leq 0\end{cases}$$

whose derivative

$$f'(x) = \begin{cases} \frac{x}{x+1} & x >0 \\ 0 &x\leq 0\end{cases} $$

is monotone increasing on $\mathbb{R}$ and hence $f(x)$ is convex.

Moreover, $f'(x)\to 1$ as $x\to +\infty$ and

$$x(f'(x)-1) = -\frac{x}{x+1} \to -1$$

as $x\to+\infty$.

Answer 2

We can make the limit anything we want. Consider $$ x(f'(x)-a)=b\tag1 $$ Solving $(1)$ gives $$ f(x)=ax+b\log(x)+c\tag2 $$ To make $f''(x)\ge0$, we need $-\frac{b}{x^2}\ge0$. That is, set $b\lt0$ in $(2)$.

If you want this to be defined for all $x\in\mathbb{R}$, then use $$ f(x)=\left\{\begin{array}{} ax+b\log(x)+c&\text{if }x\ge1\\ c-b+(a+b)x&\text{if }x\lt1 \end{array}\right.\tag3 $$