1
$\begingroup$

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


I posted this question on MSE, but have not received any satisfying answer. So I post it here.

$\endgroup$
1

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .