Does there exist a sequence of decreasing continuous functions $(f_n)_{n\in\mathbb{N}}$ satisfying the following two conditions?

  • For every $n\in\mathbb{N}$, $\lim_{x\to\infty}f_n(x)=0$;
  • For any other decreasing continuous function $g$ tending to zero at infinity, there exists $n\in\mathbb{N}$ so that $\frac{g}{f_n}$ is a decreasing to zero function.

The spirit is quite clear, it is tempting to say that we can approach the constant function equal to zero with a sequence that eventually majorizes every other function tending to zero at infinity. But, I cannot come up with such an example satisfying the second condition above and I am not even sure that this is true. Of course, any idea allowing to answer this with restrictions (for all $g$ convex or satisfying some regularity conditions,... or asking only that $\lim_{x\to\infty}\frac{g(x)}{f_n(x)}=0$ or $\frac{g}{f_n}$ just decreasing,...) is welcome.

  • 5
    $\begingroup$ Fix a sequence $(f_n)$. For any $k$ take $N_k$ such that $f_1(N_k),\dots,f_n(N_k)<1/k^2$ and let $g$ be a piecewise linear function such that $g(N_k)=1/k$. Then none of $g/f_n$ tend to zero. I don't know of any sensible conditions which would make such a sequence exist. $\endgroup$
    – Wojowu
    Jan 19, 2021 at 17:25
  • $\begingroup$ The “sequence” structure appears superfluous— it makes sense to ask whether the conditions hold for any set of functions. It seems that a countable set will not work, but one could ask whether a strictly uncountable set with cardinality less than the set of all functions from reals to reals will work. $\endgroup$
    – Vik78
    Jan 19, 2021 at 17:34
  • 4
    $\begingroup$ @Vik78: The set of continuous functions on $[0,\infty)$ which vanish at infinity has cardinality only $\mathfrak{c}$ (it is a complete separable metric space in the uniform metric), so if you want a set of lower cardinality but uncountable, you need the continuum hypothesis to fail. This seems closely related to the dominating number $\mathfrak{d}$. $\endgroup$ Jan 19, 2021 at 17:45
  • $\begingroup$ @Nate right, I missed the continuity assumption. Thanks for pointing it out. I guess the question then falls to whether there is a “nice” set of continuous functions of uncountable cardinality satisfying the criteria, or if we must relax continuity. $\endgroup$
    – Vik78
    Jan 19, 2021 at 18:29

1 Answer 1


Let $C_d$ be the space of all decreasing continuous functions tending to 0 at infinity, equipped with the sup metric $d_\infty$. Note this is a complete metric space.

Suppose such a sequence $f_n$ did exist. Then for every $g \in C_d$ there would exist $n$ such that $g/f_n$ is decreasing to 0; since $g/f_n$ is continuous, it would therefore be bounded, so there would exist $K$ such that $|g| \le K |f_n|$. So if we let $E_{K,n} = \{g \in C_d : |g| \le K |f_n|\}$, we would have $\bigcup_{k,N=0}^\infty E_{K,n} = C_d$. I claim that each $E_{K,n}$ is nowhere dense, and so this contradicts the Baire category theorem.

It is easy to see that $E_{K,n}$ is closed in $C_d$. To see it also has empty interior, fix $g \in E_{k,n}$ and $\epsilon > 0$. Since $K f_n$ vanishes at infinity, there is $M$ so large that $K |f_n(x)| < \epsilon/3$ for all $x > M$. Choose $h \in C_d$ with $\sup |h| < \epsilon$ and with $h(x_0) > 2 \epsilon/3$ for some $x_0 > M$. Then $g+h \in C_d$ with $d_{\infty}(g, g+h) < \epsilon$, but $(g+h)(x_0) > \epsilon/3$ so $g+h \notin E_{K,n}$


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