Let us say that a bounded smooth function $f:\mathbb{R}\rightarrow\mathbb{R}$ has *vanishing variation at infinity* (or satisfies "property $A$" for short) if, for any $r\neq 0$, we have


$$\lim\_{x\rightarrow\infty}\frac{|f(x+r)-f(x)|}{|r|} = 0.$$


In particular this means that


$$\lim\_{r\rightarrow 0}\left(\lim\_{x\rightarrow\infty}\frac{|f(x+r)-f(x)|}{|r|}\right) = 0.$$


An example of such a function is any smooth function whose *gradient vanishes at infinity* (or satisfies "property $B$" for short): that is,


$$\lim\_{x\rightarrow\infty}\left(\lim\_{r\rightarrow 0}\frac{|f(x+r)-f(x)|}{|r|}\right) = 0.$$


On the other hand, there are functions satisfying property $B$ that do not satisfy property $A$. One type of example is any smooth function that has a limit as $x\rightarrow\infty$ but whose derivative does not go to zero; however note that such functions can always be approximated in the uniform norm by a sequence of functions satisfying property $A$.


**Question**: can *any* function satisfying $A$ be approximated in the uniform norm by a sequence of functions satisfying $B$? More precisely: given any $f$ satisfying $A$, can we find a sequence of functions $(f\_i)$ satisfying $B$ such that for any $\epsilon > 0$, $||f-f\_i||\_\infty<\epsilon$ for all $i$ sufficiently large?



Uniformly approximating a function of vanishing variation by functions of vanishing gradient
Let $F(x) = \exp(f(\log x))$. Then $f$ has property A if and only if $$\frac{F(\lambda x)}{F(x)} = \exp(f(\log x + \log \lambda) - f(\log x)) \to \exp(0) = 1$$ as $x \to \infty$ for any $\lambda > 0$, that is, $F$ is slowly varying at infinity.


Once we realise this, we open the book by [Bingham, Goldie and Teugels](https://books.google.com/books?id=VCU5D4jMfiEC) and search for the result that we need. This time this is Theorem 1.3.1, which tells us that $f$ has property A *if and only if* $$f(x) = a(x) + \int\_{x\_0}^{x} b(y) dy$$ for some $x\_0$, some $a(x)$ which converges as $x \to \infty$ and some $b(y)$ which converges to zero as $y \to \infty$.


Let $A$ be the limit of $a(x)$ as $x \to \infty$, and consider $$g(x) = A + \int\_{x\_0}^x b(y) dy.$$ Then $g$ has property B and $f(x) - g(x)$ converges to zero as $x \to \infty$. This is essentially what we want: simply define $$f\_n(x) = \phi(x - n) g(x) + (1 - \phi(x - n)) f(x),$$ where $\phi$ is smooth, $\phi = 0$ on $(-\infty, 0]$, $\phi = 1$ on $[1, \infty)$ and $0 \leqslant \phi \leqslant 1$. It is then easy to see that $f\_n$ has property B, and $\|f - f\_n\|\_\infty \le \sup\{|f(x) - g(x)| : x \geqslant n\} \to 0$ as $n \to \infty$.




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Edit: If, as suggested in the comment below, we assume that $\operatorname{diam} f(B(x, 1)) \to 0$ as $x \to \infty$, then the solution is much simpler (here $B(x, 1)$ is the unit ball centred at $x$). Consider a smooth function $\phi$ supported in the unit ball and with total mass $1$, and consider $g = f \* \phi$. Then $$ |g(x) - f(x)| = \left|\int\_{B(x, 1)} (f(y) - f(x)) \phi(y - x) dy\right| \leqslant \operatorname{diam} f(B(x, 1)) \|\phi\|\_1 $$ and $$\begin{aligned}|g'(x)| & = \left|\int\_{(x, 1)} f(y) \phi'(y - x) dy\right| \\ & = \left|\int\_{(x, 1)} (f(y) - f(x)) \phi'(y - x) dy\right| \leqslant \operatorname{diam} f(B(x, 1)) \|\phi'\|\_1 . \end{aligned}$$ Therefore, defining $f\_n$ as in the original answer, we get the desired result.


This answer works in higher dimensions as well. The difficult part in the original answer was getting a uniform limit in $r$, which is a variant of the Karamata–Korevaar *uniform convergence theorem*. (I did not check if there are higher-dimensional analogues of this result).


