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?


1 Answer 1


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 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 $n \to \infty$.

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).

  • $\begingroup$ Does this equivalence also generalise to functions $f:\mathbb{R}^2\rightarrow\mathbb{R}$? $\endgroup$
    – geometricK
    Jan 18, 2018 at 2:41
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    $\begingroup$ @ougoah: I think there is more than one way one could state the corresponding multi-dimensional property (is $|x| \to \infty$ or $x_j \to \infty$ for all $j$? is $x$ in an orthant or in a hyperplane?). In any case I would start by reviewing literature on multivariate regularly varying functions (which I do not know) in search of a similar result. $\endgroup$ Jan 18, 2018 at 6:49
  • $\begingroup$ Thanks for the suggestion. I was thinking of restating it in terms of vanishing variation outside compact sets. That is, $f$ satisfies property $A$ if for any $r>0$, for any $\epsilon > 0$ there exists a compact $K$ such that for all $x\in\mathbb{R}^2\backslash K$, $\text{diam} f(B_x(r)) < \epsilon$. $\endgroup$
    – geometricK
    Jan 18, 2018 at 11:30
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    $\begingroup$ Oh, this is quite simple, but too long for a comment. I will update the answer in a couple of minutes. $\endgroup$ Jan 18, 2018 at 12:29
  • $\begingroup$ I see! Do you see a way to do it on a general Riemannian manifold instead of $\mathbb{R}^2$? I assume to do so would require replacing convolution by something else. Perhaps this is hard... $\endgroup$
    – geometricK
    Jan 18, 2018 at 18:38

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