Does an isometric automorphism of $L_p (\mu, X)$ preserve pointwise convergence?

Question

Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Here we use the Bochner integral. Let $p, q \in (1, \infty)$ such that $p^{-1}+q^{-1}=1$. In an attempt to formalize the ideas in this comment, I have come across below questions, i.e.,

1. Let $\psi$ is an isometric automorphism of $L_{p}(\mu, X)$. Let $f, f_n \in L_{p}(\mu, X)$ such that $f_n \to f$ pointwise $\mu$-a.e. Does $\psi(f_n) \to \psi(f)$ pointwise $\mu$-a.e.?

2. Let $X^{*}$ have the Radon-Nikodým property with respect to $\mu$. Then there is an isometric isomorphism $\varphi:L_{p}(\mu, X)^* \to L_{q} (\mu, X^*)$. Let $H,H_n \in L_{p}(\mu, X)^*$ such that $H_n \to H$ pointwise. Does $\varphi(H_n) \to \varphi(H)$ pointwise $\mu$-a.e.?

I feel that isometric isomorphisms do not preserve pointwise convergence. However, I could not come up with a counter-example. Could you elaborate on my questions?

Answer 1

If the convergence of $f_n$ is also in $L_p$ then you can extract a subsequence converging pointwise for the isometric sequence. Wlog, we can assume $f_n \rightarrow 0$ pointwise $\mu-a.e.$. Assume $|f_n(x)| \leq |g(x)|$ with $g \in L_p$, then $f_n \rightarrow 0$ in $L_p$. Hence $\psi(f_{n}) \rightarrow 0$ in $L_p$. Hence there is a subsequence $\psi(f_{n_i}) \rightarrow 0$ pointwise $\mu$-a.e.

Assume $|f_n(x)| \leq |g(x)|$ with $g \in L_p$: as $\psi$ is an isometry: $$\int |\psi(f_n)|^p = \int |f_n|^p$$ $$\int I_{E_n}|\psi(f_n)|^p \leq \int |f_n|^p$$ where $E_n = \{x: |\psi(f_n)| \geq \epsilon\}$. $$\epsilon^p \mu(E_n) \leq \int |f_n|^p$$ taking limit on both sides: $$\epsilon^p \lim_{n \rightarrow \infty} \mu(E_n) \leq \lim_{n \rightarrow \infty} \int |f_n|^p = \int \lim_{n \rightarrow \infty} |f_n|^p = 0$$ Hence for every $\epsilon>0$ we have $\lim_{n \rightarrow \infty} \mu(E_n) = 0$, we now need to prove $ \mu(\cup_{n \geq k} E_n) \rightarrow 0$. See if u can get this under some conditions. For example, $ \mu(\cup_{n \geq k} E_n) \rightarrow 0$ as $k \rightarrow \infty$ can be proved by assuming $\lim_{k \rightarrow \infty} \sum_{n \geq k} ||f_n||_p = 0$. So atleast $\lim_{k \rightarrow \infty} \sum_{n \geq k} ||f_n||_p = 0$ implies pointwise convergence of $\psi(f_n)$ to $0$.