When I read the derivation of PDE for a lookback option, it makes use the following result:
$J_{nt}=\left(\frac{1}{t}\int_{0}^{t} S_{s}^{n} ds\right)^{\frac{1}{n}}$ and $\lim_{n\rightarrow \infty} J_{nt}=\max_{0\leq s\leq t}S_{s}$. I would like to ask how to prove the results? Assume that $S$ is a continuous function. Also, I would like to ask how to define $J_{nt}$ such that it tends to $\min_{0\leq s\leq t}S_{s}$?
The first result is a special case of the following: $||f||_p \to ||f||_\infty $as $p \to \infty$ for any $L^{\infty}$ function $f$ provided we are integrating with respect to a probability measure. I don't think you can get the minimum of $S_s$ by this kind of limit. Proof of the result claimed: Obviously $\int |f|^{p} d\mu \leq ||f||_{\infty} ^{p}$ so wee that $limsup ||f||_p \leq ||f||_\infty$. Now let $\epsilon >0$ and consider $E=\{x:|f(x) | > ||f||_\infty -\epsilon\}$ By definition of $||f||_\infty$ the set E has positive measure. Now $\int |f|^{p} d\mu \geq\int_E |f|^{p} d\mu \geq (||f||_\infty-\epsilon)^{p} m(E)$. Take $p-th$ root on both sides and note that $(m(E))^{1/p} \to 1$ as $p \to \infty$.