Conceptual explanation of geometric mean as a limit of power means

Question

Let $x_1,\dots,x_n$ be positive real numbers and $p\in\mathbb{R} -\{0\}$. The power mean $M_p(x_1,\dots,x_n)$ of exponent $p$ is defined by $$ M_p(x_1,\dots,x_n)=\left( \frac 1n\sum_{i=1}^n x_i^p \right)^{1/p}. $$ By taking logarithms and applying L'Hôpital's rule (or just the definition of a derivative), we get $$ \lim_{p\to 0} M_p(x_1,\dots,x_n) = \sqrt[n]{x_1\cdots x_n}, $$ the geometric mean of $x_1,\dots,x_n$. Thus the "correct" definition of $M_0$ is $M_0(x_1,\dots,x_n) = \sqrt[n]{x_1\cdots x_n}$. All this is well known, but I am wondering if there is some conceptual explanation, not involving computation, for the value of $M_0(x_1,\dots,x_n)$.

Answer 1

$(\frac{a^{1/n} + b^{1/n}}{2})^n$ has normalized symmetric binomial (i.e., close to normal) distribution of coefficients convolved with some bounded expressions in $a$ and $b$ (of the form $a^\theta b^{1 - \theta}$). The normal distribution is sharply peaked around its mean so you end up with the middle terms with $\theta = 1/2$ in the limit of large $n$. This generalizes to more variables and weights.

That is still a calculation, but one where the answer is easy to foresee at the outset and the idea can be explained in a few words.

Answer 2

A reasonably intuitive way to "see" that the limit must be the geometric mean is the plausible and useful observation that any power mean can be expressed in terms of the midpoint means, $M_p(x,y)=((x^p+y^p)/2)^{1/p}$, recursively if the number of variables is not a power of 2. See my answer to a related question.

Simple algebra then proves that for $n=2$, and all $p\neq0$, $M_p M_{-p}=x_1x_2$ so letting $p\rightarrow0$ we have $M_0^2=x_1x_2$ and the general limit value follows immediately.

Note that this doesn't use calculus, transcendental functions, or in fact anything other than the power means themselves and their continuity.

Update:

A further low tech way to see the result uses functional equations.

We simply note that the power means satisfy $$M_{rp}(x_1,\cdots,x_n)=M_p(x_1^r,\cdots,x_n^r)^{1/r}.$$

Setting $p=0$ gives

$$M_{0}(x_1,\cdots,x_n)=M_0(x_1^r,\cdots,x_n^r)^{1/r}$$ for all $r\in \mathbb{R}-\{0\}.$

Then it is intuitively clear since $M$ is symmetric and $M_p(\lambda \mathbb x)=\lambda M_p(\mathbb x)$ that $M_0$ must be the geometric mean.

You can prove this formally by induction starting with $n=2$. Let $f(x)=M_0(x,1)=f(x^r)^{1/r}$, by the above. Then setting $x=e$, $r=\log X$ we have $f(X)=f(e)^{\log X}=X^{\log f(e)}=x^\mu$ where $\mu$ is constant.

Hence $M_0(x,y)=yM_0(x/y,1)=yf(x/y)=y(x/y)^\mu=y^{1-\mu}x^{\mu}$. SInce $M_0$ is symmetric in $x$ and $y$ we have $\mu=1/2$ and $M_0(x,y)=x^{1/2}y^{1/2}$. The other cases $n>2$ follow in a similar manner.

Further update:

Actually probably the most intuitive way is just to note, as Iosif did as well, that AM-GM or Jensen's inequality tells you $M_p\geq GM\geq M_{-p}$. Then just take the limit as $p\rightarrow 0$.

Answer 3

$\newcommand\o\overline$ This proof uses only the arithmetic-geometric mean (AGM) inequality and the fact that for any smooth even function $g\colon\mathbb R\to\mathbb R$ we have $g'(0)=0$.

To simplify the writing, for any function $f\colon\mathbb R\to\mathbb R$ let $$\o{f(x)}:=\frac1n\,\sum_1^n f(x_i).$$

We have to show that $$M_p:=(\o{x^p})^{1/p}\to M_0:=\exp\,\o{\ln x}$$ as $p\to0$.

Take any real $p>0$. Replacing the $x_i$'s in the AGM inequality $$\o x\ge \exp\,\o{\ln x} \tag{1}$$ by the $x_i^p$'s, we have $M_p\ge M_0$. Similarly, replacing the $x_i$'s in (1) by the $x_i^{-p}$'s, we have $M_{-p}\le M_0$. So, $$M_{-p}\le M_0\le M_p.$$ It remains to show that $M_p/M_{-p}\to1$ as $p\downarrow0$ or, equivalently, that $$g(p):=\ln\o{x^p}+\ln\o{x^{-p}}=o(p),$$ which follows because the function $g$ is smooth and even, with $g(0)=0$. $\Box$

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The expression $\exp\,\o{\ln x}\,[=(x_1\cdots x_n)^{1/n}]$ for the geometric mean arises naturally, as an instance, with $f=\exp$, of the more general mean of the form $f\big(\o{f^{-1}(x)}\big)$ with a continuous increasing function $f$. So, the geometric mean is just a logarithmically/exponentially re-scaled version of the arithmetic mean. Also, the AGM inequality (1) is an instance of Jensen's inequality for the concave function $\ln$ or, equivalently, for the the convex function $\exp$.

Answer 4

I am not sure if the following is a conceptual explanation rather than a calculation:

Writing $x_i=e^{u_i}$ and letting $p\to0$, we have $$M_p=\Big(\frac1n\,\sum_1^n e^{pu_i}\Big)^{1/p} =\Big(1+\frac p{n+o(1)}\,\sum_1^n u_i\Big)^{1/p}\to\exp\Big(\frac1n\,\sum_1^n u_i\Big)=M_0,$$ where $M_r:=M_r(x_1,\dots,x_n)$.

(I guess in any case we need to show that $M_p\to M_0$ as $p\to0$. Here, at least we do not explicitly use the l'Hospital rule or differentiation.)