# Continuous weighted means

Let $f$ and $\omega$ be a positive functions over $(a,b)$ ( $f:(a,b)\to\mathbb R^+$, and $\omega:(a,b)\to\mathbb R^{\ge0}$ ) such that $$\int_a^b\omega(x)dx=1$$

Under which conditions can I guaranty that: $$\lim_{t\to0^+}\left(\int_a^b\omega(x)f^t(x)dx\right)^{\frac1t}=\exp\int_a^b\omega(x)\ln f(x)dx$$ and $$\lim_{t\to0^-}\left(\int_a^b\omega(x)f^t(x)dx\right)^{\frac1t}=\exp\int_a^b\omega(x)\ln f(x)dx$$

A second conjecture is that $$\lim_{t\to+\infty}\left(\int_a^b\omega(x)f^t(x)dx\right)^{\frac1t}=\sup_{x\in(a,b)} f(x)$$ provided that $\omega(x_s)\ne0$ for $f(x_s)=\sup_{x\in(a,b)} f(x).$ Analogously $$\lim_{t\to-\infty}\left(\int_a^b\omega(x)f^t(x)dx\right)^{\frac1t}=\inf_{x\in(a,b)} f(x)$$

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