A proof is sought.
Asset $A$ is a standard or benchmark investment. Asset $B$ is a prospective investment. Usually investors want to know the historical rate of return of an investment. Suppose it is a preference to pay attention to the difference between the historical rates of return of these two investments.
$p_{at}$ is the price of asset $A$ at time $t$.
$p_{bt}$ is the price of asset $B$ at time $t$.
$t$ denotes a quarter year, so $p_{xx}$ is available on December 31, March 31, June 30, September 30, etcetera. The rates of return are produced as follows.
$$a_1=\frac{p_{a1}-p_{a0}}{p_{a0}}=\frac{p_{a1}}{p_{a0}}-1$$ $$b_1=\frac{p_{b1}-p_{b0}}{p_{b0}}=\frac{p_{b1}}{p_{b0}}-1$$
Let $a$ and $b$ be the geometric means of many quarterly rates. Let $m$ denote some large integer.
$$a=\Biggl[\prod_1^m{(1+a_n)}\Biggr]^{(1/m)}-1$$ $$b=\Biggl[\prod_1^m{(1+b_n)}\Biggr]^{(1/m)}-1$$
Let $d_{annualized}$ denote the difference of the annualized geometric mean rates. Since $t$ advances $4$ times per year $n=4$ in the following.
$$annualized(a)=(1+a)^n-1$$ $$annualized(b)=(1+b)^n-1$$ $$d_{annualized}=annualized(b)-annualized(a)$$ $$d_{annualized}=[(1+b)^n-1]-[(1+a)^n-1]$$ $$d_{annualized}=(1+b)^n-(1+a)^n$$
Let $d$ denote the difference in the geometric mean of quarterly rates.
$$d=b-a$$
Via trials (number substitution) it was observed that the annualization of the difference of the geometric means of quarterly rates does not in general equal the difference of the annualized rates. The exception is when $n=1$ but usually $n \ne 1$ and in this example $n=4$.
$$annualized(d) \ne d_{annualized}$$ $$(1+d)^n-1 \ne d_{annualized}$$
When $n \ne 1$ does there exist an alternative function $f(d)$ that equals $d_{annualized}$ or a proof of nonexistence of such a function? A hypothetical practical application might be the situation wherein software has access to $d$ but does not have access to $a$ and $b$ and there is still the goal of calculating the value $d_{annualized}$.
$$f(d)=\quad?\quad =d_{annualized}$$
You would like to have a function $f$ such that for a fixed $n>1$ and all $a, b\in\mathbb{R}$: $$f(b-a)= f(d)=d_{annualized}=(1 + b)^n - (1 + a)^n .$$
This is not possible. The easiest way to see this, is to observe that there are numbers $x,y$ and $\tilde{x},\tilde{y}$ such that $$ y - x = \tilde{y} - \tilde{x}$$ but $$ (1 + y)^n - (1 + x)^n \neq (1 + \tilde{y})^n - (1 + \tilde{x})^n.$$
Take for example $n=4$ with $y=0.2$, $x=0.1$ and $\tilde{y}=0.4$, $\tilde{x}=0.3$. Then $$ (1 + 0.2)^4 - (1 + 0.1)^4 = 0.6095$$ but $$ (1 + 0.4)^4 - (1 + 0.3)^4 = 0.9855.$$