Let me given with an obvious example. Let $\Omega\subset{\mathbb R}^n$ be an open domain. If $f,g\in L^1(\Omega)$ and $f,g\ge0$, then $\sqrt{fg}\,\in L^1(\Omega)$.
Now let me replace the absolutely continuous measures $f(x)dx$ and $g(x)dx$, by a pair $\lambda,\mu$ of non-negative bounded measures on $\Omega$.
Is there a natural way to define a geometric mean $\sqrt{\lambda\mu}$ as a non-negative measure ?
Here is an incomplete attempt: remark that if $a,b\in[0,\infty)$, then $\sqrt{ab}=\inf_{p,q>0}\frac{ap+bq}{2\sqrt{pq}}\,$. This suggests to define $$\langle\sqrt{\lambda\mu},\phi\rangle:=\inf_h\{\langle\lambda,h\phi\rangle+\langle\mu,\frac\phi{h}\rangle\}\qquad\forall\phi\in C^+_K(\Omega)$$ where the infimum runs over function $h\in C(\overline{\Omega})$ that are strictly positive. This definition has several nice features: - it yields the correct function when applied to $L^1$-functions, - as a function of $\phi$, it is positive and homogeneous of degree one. However, it is not clear whether this defines a linear form.
