Prove that the curve of the function $ f $ has an asymptote near infinity which is parallel to the line $ y=x $

Let $ f $ be a continuous function from $ \Bbb R $ to $ \Bbb R $ satisfying the condition $$(\forall x,y\in\Bbb R) \;|f(x)-f(y)|\ge |x-y|$$

If we assume that $ f $ is increasing and that $$(\forall x\in \Bbb R)\;f(x)<x,$$ Prove that the curve $ C_f $ has an asymptot parallel to the line $ y= x, $ near $ +\infty$.

I tried to prove that there exists a real constant $ a$ such that $$\lim_{x\to+\infty}(f(x)-x)=a $$ but i couldn't. Any help will be appreciated.

Let $g(x) = x - f(x)$. Then $g(x) > 0$. We show that $g(x)$ is monotone decreasing, which implies $\lim_{x \to \infty} g(x)$ exists, by the (continuous) monotone convergence theorem.

Suppose $y > x$. Then, \begin{align*} g(y) - g(x) &= (y - x) - (f(y) - f(x)) \\ &= (y - x) - |f(y) - f(x)| & \text{($f$ is increasing)} \\ &\le (y - x) - |y - x| \\ &\le 0, \end{align*}

as required.

Partial answer

From the first condition, taking $0=y \lt x$ we get as $f$ is supposed to be increasing

$$\frac{f(x)}{x} -\frac{f(0)}{x} \ge 1$$ hence $$\liminf\limits_{x \to \infty} \frac{f(x)}{x} \ge 1.$$ The inequality $f(x) \lt x$ leads to $$\limsup\limits_{x \to \infty} \frac{f(x)}{x} \le 1.$$ Therefore $$\liminf\limits_{x \to \infty} \frac{f(x)}{x} = \limsup\limits_{x \to \infty} \frac{f(x)}{x} =1$$ proving that $\lim\limits_{x \to \infty} \frac{f(x)}{x}$ exists and is equal to $1$.

To get the conclusion, we can try to prove that if $f(x)-x$ has two limit points, then those have to be equal.