How to prove the Hubble law is the unique expansion law compatible with homogeneity and isotropy?

Question

In the book physical foundations of cosmology, it says that Hubble's Law is unique and a problem seems to be a hint of proving that.

In order for a general expansion law,v=f(r,t), to be the same for all observers, the function f must satisfy the relation $$f(\bf{r_{CA}}−\bf{r_{BA}},t) = f(\bf{r_{CA}},t)−f(\bf{r_{BA}},t),$$ where ABC are three points in space. Show that the only solution of this equation is given by the Hubble law.

With a little help from a Taylor approximation, I can convince myself that $f$ should be a linear function without a constant. But it seems to me that this is not good enough for a proof. How can one prove it in a more mathematical way? Thanks!

Answer 1

First, since $t$ does not enter into the given relation we suppress the time-dependence of $f$. Additionally, while the problem presumably assumes ordinary 3D space I'll consider $f$ as a function $\mathbb{R}^n\to \mathbb{R}$ (i.e., $n$-dimensional space). Now let $\mathbf{x}=\mathbf{r}_{CA}-\mathbf{r}_{BA}=\mathbf{r}_{CB}$, $\mathbf{y}=\mathbf{r}_{BA}$ and rearrange the relation as $$f(\mathbf{x})+f(\mathbf{y})=f(\mathbf{x}+\mathbf{y})$$ Since the points $A,B,C$ are arbitrary, the above relation should likewise be valid for arbitrary real $\mathbf{x},\mathbf{y}$. It is easy to see that the $f(\mathbf{x}):=\mathbf{c}\cdot \mathbf{x}=\sum_{i=1}^n c_i x_i$ is a solution for any $\mathbf{c}\in\mathbb{R}^n$ which I take to be the desired "Hubble law". Is this the only solution?

This is a multidimensional version of Cauchy's (additive) functional equation. Via the MathOverflow question here, the multidimensional version (assuming continuous $f$) appears in Kuczma's An Introduction to the Theory of Functional Equations and Inequalities. Quoting from the answer to the aforementioned Q&A: "[Theorem 5.52] shows that the only continuous solutions are precisely the functions of the form $\mathbb{R}^n\to \mathbb{R}:(x_1,\ldots,x_n)\mapsto \sum_{i=1}^n c_i x_i$ with $c_1,\ldots,c_n\in\mathbb{R}$." The proof is a straightforward extension of the one-variable version; this has appeared on MSE many times (see for instance Martin Sleziak's summary here) so I won't consider this further.

What if we drop the condition of continuity? In that case there do exist nontrivial solutions to Cauchy's additive functional equation. However, these solutions are well-known to be quite pathological; in particular, such functions are known to not be measurable. For a summary of basic results on Cauchy's functional equation in the 1D case, see Martin Sleziak's summary as referenced above; for more definitive results in the multidimensional case, Kuczma seems the appropriate reference.

Answer 2

I spent a lot of time reading about Hubble's Law only to find that in a math context this problem isn't too hard to explain.

This function says that the change in $r$ corresponds to a constant change in $f$. So, let $x=\frac{\partial f}{\partial r}+\frac{\partial f}{\partial t}\frac{dt}{dr}$. Since we assume that $t$ is a constantly changing function, it's derivative with respect to $r$ is 0. this means that $x=\frac{\partial f}{\partial r}$. We can integrate this function with respect to $r$ \begin{gather} \int \frac{\partial f}{\partial r}\,dr=\int x\,dr = xr + c \end{gather}

I am unsure of as to why $c$ would be set to 0. There's probably an argument in there somewhere but I'm physics'd out for the day.