Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Here we use the [Bochner integral](https://math.stackexchange.com/questions/4298588/dominated-convergence-theorem-for-banach-space). Let $p \in [1, \infty)$ and $q \in (1, \infty]$ such that $p^{-1}+q^{-1}=1$. In an attempt to formalize the ideas in [this comment](https://math.stackexchange.com/questions/4577980/let-h-h-n-in-l-p-such-that-h-n-f-hf1-x-n-for-all-f-in-l-p-a#comment9636279_4577980), I have come across below questions, i.e.,


1. Let $\psi$ be an isometric automorphism of $L\_{p}(X,\mu, E)$. Let $f, f\_n \in L\_{p}(X,\mu, E)$ 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](https://math.stackexchange.com/questions/4546633/construct-conditional-expectation-on-banach-spaces-with-radon-nikod%c3%bdm-property) with respect to $\mu$. Then there is a canonical isometric isomorphism $\varphi:L\_{p}(X,\mu, E)^\* \to L\_{q} (X,\mu, E^\*)$ such that
$$
K (f) = \int\_\Omega \langle \varphi(K), f \rangle \mathrm d \mu \quad \forall K \in L\_{p}(X,\mu, E)^\*, \forall f \in L\_{p}(X,\mu, E)
$$
Let $H,H\_n \in L\_{p}(X,\mu, E)^\*$ 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?




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I posted [this question](https://math.stackexchange.com/questions/4580930/does-an-isometric-automorphism-of-l-p-mu-x-preserve-pointwise-convergence) on MSE, but have not received any satisfying answer. So I post it here.



Does an isometric automorphism of $L_p (X,\mu, E)$ preserve pointwise convergence?
Point 1) does not hold for the Fourier transform on $L^2({\bf R}, {\bf C})$, which is an isometry for a well chosen normalization.


Consider the sequence $$f\_n(x) = n {\bf 1}\_{[-1/n,1/n]}(x)$$ which converges to 0 for all $x\neq 0$. The Fourier transform of $f\_n$ is proportional to $${sin(x/n) \over {x/n}}$$ which converges to $1$ as $n$ goes to infinity for all $x \neq 0$. And the constant function 1 is not the Fourier transform of 0.


