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Suppose $w \in L^{1}(G)$ and $w \geq 0$. The geometric mean of $w$ is defined by $$ \Delta(w) = \exp \int_{G} \log w(x) \ dx$$

where $\Delta(w) = 0$ if the integral is $-\infty$. What is the intuition behind this definition?

Source: Fourier Analysis on Groups by Walter Rudin

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1 Answer 1

up vote 5 down vote accepted

Isn't this an extension of the geometric mean in discrete setting?

If $a$ is the geometric mean of $a_1,a_2,a_3,\ldots,a_n$ then $$a=\sqrt[n]{a_1 \times a_2 \times \cdots \times a_n}$$ which implies $$\log(a) = \frac{\log(a_1) + \log(a_2) + \cdots + \log(a_n)}{n}$$ i.e. $\log(a)$ is the arithmetic mean of $\log(a_i)$'s.

When we talk of average in continuous sense, we replace the summation by integrals and hence

$$\displaystyle \log(\Delta(w)) = \frac{\int_G \log(w(x)) dx}{\int_G dx}$$

Hence,

$$\Delta(w) = \exp \left(\frac{\int_G \log(w(x)) dx}{\int_G dx} \right)$$

and if $l \rightarrow -\infty$, then $\displaystyle \lim_{l \rightarrow -\infty} e^l = 0$ and hence $\Delta(w) = 0$ if the integral is $-\infty$

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