# A subadditive bijection on the positive reals

Question. Does there exist a subadditive bijection $$f$$ of the positive reals $$(0,\infty)$$ such that $$\liminf_{x\to 0^+}f(x)=0 \,\,\,\text{ and }\,\,\,\limsup_{x\to 0^+}f(x)=1\,?$$

Ps1. I guess the answer is no. However, it is affirmative if we replace "bijective" by "injective" by the following example: $$f(x)=x$$ if $$x$$ is rational, otherwise $$f(x)=x+1$$.

Ps2. The answer is also affirmative if we replace $$\limsup_{x\to 0^+}f(x)=1$$ with $$\limsup_{x\to 0^+}f(x)\neq 0.$$ Indeed, let $$(x_i: i \in I)$$ be an Hamel basis of the vector space $$\mathbf{R}$$ over $$\mathbf{Q}$$. Fix a nontrivial permutation $$\sigma$$ of $$I$$. For each $$r=\sum_j \lambda_jx_j>0$$ define $$f(r)=|g(r)|$$, where $$g(r)=\sum_j \lambda_j x_{\sigma(j)}$$ (thanks i707107). Note that $$g$$ is an additive bijection, hence $$f(x+y)=|g(x)+g(y)| \le f(x)+f(y)$$ for all $$x,y>0$$, so $$f$$ is subadditive. Now suppose $$f(x)=f(y)$$ with $$x,y>0$$. Then either $$g(x)=g(y)$$ so $$x=y$$, or $$g(x)=-g(y)$$ so each coefficient of $$\lambda_j$$ (in $$x$$) is the inverse of the corresponding (in $$y$$), so $$x=-y$$: if $$x,y>0$$, then $$f(x)=f(y)$$ implies $$x=y$$ so $$f$$ is injective.

About surjectivity, fix $$x=\sum_{j} \lambda_j x_j$$ and define $$y=\sum_j \lambda_j x_{\sigma^{-1}(j)}$$. Hence $$z:=\max(y,-y)>0$$ and $$f(z)=x$$; so $$f$$ is subjective.

Finally, it is known that $$g$$ has a graph dense in $$\mathbf{R}^2$$, from which it holds also $$\liminf_{x\to 0^+}f(x)=0$$ and $$\limsup_{x\to 0^+}f(x)=\infty$$.

• I realized the example is wrong. However, I write it since I don't know if it can be fixed: let $(x_i)_{i \in I}$ be an Hamel basis of $\mathbf{R}/\mathbf{Q}$ and let $\sigma$ be a nontrivial permutation of $I$. Then, if $r>0$ has representation as finite sum $\sum_j \lambda_jx_j$, we define the additive function $h(r)=\sum_j |\sigma(\lambda_j)|x_j$. However, I realized now that, although $h(r)>0$ for each $r$, the function is neither injective not surijective. (I remove that comment.) Commented Jun 26, 2018 at 10:18
• What if $h(r)=|\sum_j \lambda_j x_{\sigma(j)}|$ ? Commented Jun 26, 2018 at 16:49
• I think the above one is subadditive bijection with $\liminf_{x\rightarrow 0+} h(x)=0$, and $\limsup_{x\rightarrow 0+} h(x)=\infty$. Commented Jun 26, 2018 at 19:03
• You are correct. I added the details in the OP Commented Jun 26, 2018 at 20:26
• Oh sorry, I forgot that we want $f$ to map $(0,\infty)$ onto $(0,\infty)$... So I will delete my useless comment. Commented May 22, 2022 at 20:38

Using ideas described in Gowers blog post about Zorn's Lemma we will define a subadditive function $$f\colon ℝ_+ → ℝ_+$$ with the desired property.

Let $$\mathcal{B} = \{e_β\}_{β ∈ B}$$ be a Hamel Basis of $$ℝ$$ over the field of rationals. Since $$\mathcal{B}$$ remains a Hamel Basis when we replace $$e_β$$ with $$q\cdot e_β$$ where $$q$$ is non-zero rational, we can assume that all $$e_β > 0$$ and there is $$\{ e_n \}_{n \in ℤ_+} \subset \mathcal{B}$$ such that $$2^{-(n+1)}< e_n<2^{-n}$$.

For each $$n \in ℤ_+$$ pick $$a_n,b_n \in \mathbb{Q}_+$$ such that $$1 - 2^{-n} < a_n e_n < b_n e_n < 1$$. Note $$a_n e_n → 1$$. Let $$\text{span}(e_n) = \{ q e_n \mid q \in \mathbb{Q}_+\}$$ and define a piecewise linear function $$f_n \colon \text{span}(e_n) \to \text{span}(e_n)$$ passing thru there points $$(0,0), (e_n, a_n e_n)$$ and $$(b_n e_n, b_n e_n)$$: $$f_n(x) = \begin{cases} a_n x, & 0 < x ≤ e_n,\\ \frac{b_n-a_n}{b_n-1}x + e_n \frac{b_n(a_n-1)}{b_n-1}, & e_n < x ≤ b_n e_n,\\ x, & b_n e_n < x. \end{cases}$$ (Pedantical note: Point $$(0,0)$$ does not belong to the graph of $$f_n$$, but this framing makes the idea behind $$f_n$$ clear.) To familiarize with $$f_n$$'s here is graph of $$f_3$$:

Just keep in mind that domain and codomain are not $$ℝ_+$$ but $$\text{span}(e_3)$$.

Note that $$f_n$$ is well-defined since we always multiply $$e_n$$ by some positive rational, so the codomain is indeed $$\text{span}(e_n)$$. It's a bijection. Moreover, $$f_n$$ is subadditive. We verify this by boring casework at the end of this proof, but the upshot is this: In each case, we either use the fact that $$f_n|_{(0,b_ne_n]}$$ is concave and positive and hence subadditive, or that the graph of $$f_n$$ is above the identity, that is $$x \leq f_n(x)$$.

Finally, we can define $$f\colon ℝ_+ → ℝ_+$$ as $$f(x) = \begin{cases} f_n(x), & x \in \text{span}(e_n),\\ x, & \text{otherwise.} \end{cases}$$ This is a well-defined function because each $$x \in ℝ_+$$ has a unique representation in terms of Hamel Base $$\mathcal{B}$$.

Function $$f$$ is a bijection. Clearly, $$\liminf_{x → 0_+} f(x) = 0$$ since it acts like the identity function for almost all inputs, and otherwise $$f_n ≥ 0$$. To find $$\limsup$$, first note that on $$(0, 1]$$, $$f$$ is bounded from above by $$1$$. Secondly, $$e_n → 0$$ and $$1>f(e_n) > a_n e_n > 1-2^{-n}$$, so $$f(e_n) \to 1$$. Hence, $$\limsup_{x\to 0^+} f(x) = 1$$ as required.

It remains to show that $$f$$ is subadditive. Take any $$x,y \in ℝ_+$$.

• If $$x,y \in \text{span}(e_n)$$ for some $$n$$. Then $$x+y \in \text{span}(e_n)$$ and subaddivity of $$f$$ follows from the subadditivity of $$f_n$$.

• If $$x \in \text{span}(e_n)$$ for some $$n$$ but $$y \not\in \text{span}(e_n)$$ (or vice versa). Then $$f(x+y) = x+y$$. Because $$f_n(x) ≥ x$$ as well as $$f(y) ≥ y$$ we have $$f(x) + f(y) = f_n(x) + f(y) ≥ x + y = f(x+y)$$.

• If none of the above cases are true, then $$f(x+y) = x+ y = f(x) + f(y)$$.

We show that the function $$f_n$$ is subadditive. Take $$x,y \in \text{span}(e_n)$$. WLOG assume that $$x \leq y$$. If $$x,y$$ and $$x+y$$ are in one of three intervals $$(0, e_n]$$, $$[e_n, b_n e_n]$$, or $$[b_n e_n, \infty)$$, then the condition holds, since a restriction of $$f_n$$ to each of those intervals is additive. It remains to show what happens in "mixed" cases.

When $$x,y≤ e_n$$ and:

• (A1) $$e_n < x+y ≤b_n e_n$$.
• (A2) $$b_n e_n < x+y$$.

When $$x≤e_n and:

• (B1) $$y,x+y ≤ b_ne_n$$.
• (B2) $$y≤ b_ne_n < x+y$$.
• (B3) $$b_ne_n < y, x+y$$.

When (C) $$e_n < x,y ≤ b_n e_n < x+y$$.

When (D) $$e_n < x .

Note that $$f_n|_{(0, b_ne_n]}$$ is concave and positive. Hence subadditive. Thus subadditivity for cases A1 and B1 is proven. Now assume A2 holds. Then $$f_n(x+y) = x+ y ≤ f_n(x)+f_n(y),$$ since $$f_n$$ is on or above identity. The same reasoning will work for the rest of the cases.