# Sequence of tending to zero functions that majorizes any other tending to zero function

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.

• 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. Jan 19 at 17:25
• 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. Jan 19 at 17:34
• @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 $\mathfrak{d}$. Jan 19 at 17:45
• @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. Jan 19 at 18:29

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 $$g+h \notin E_{K,n}$$