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Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Fix $p \in [1, \infty)$. Let $L_p := L_p (X, \mu, E)$ and $\|\cdot\|_{L_p}$ be its norm. Here we use the Bochner integral. Let $(X_n)$ a countable measurable partition of $X$ such that $\mu(X_n) < \infty$. Let $L_p^* := L_p (X, \mu, E)^*$ and $\|\cdot\|_{L^*_p}$ be its norm. Let $H \in L_p^*$. We define $H_n \in L_p^*$ by $$ H_n (f) := H(f1_{X_n}) \quad \forall f \in L_p. $$

By dominated convergence theorem $\| f1_{\cup_{i=1}^n X_i} - f\|_{L_p} \to 0$ and thus $\sum_{i=1}^n H_i \to H$ weakly as $n \to \infty$. I feel it's not possible to have convergence in norm, i.e., $$ \bigg \|\sum_{i=1}^n H_i -H \bigg \|_{L^*_p} \not\to 0 \quad \text{as} \quad n \to \infty. $$

Could you confirm if my understanding is correct? Is there some relation between the norms of $H, H_n$?

Akira
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  • I think if $H\in L^{p'}(X,E^*)$ then you will have the convergence for $p>1$ by dominated convergence theorem. – daw Nov 16 '22 at 18:14
  • If $p=1$ you cannot have convergence in norm in general, already for $E=\mathbb R$. – daw Nov 16 '22 at 18:15
  • @daw Assume $p \in (1, \infty)$ and $X^{}$ has the Radon-Nikodým property with respect to $\mu$. Then there is an isometric isomorphism $\varphi:L_{p}(\mu, X)^ \to L_{q} (\mu, X^)$ where $p^{-1}+q^{-1}=1$. Let $$ K_n := \sum_{i=1}^n H_i \quad \forall n \in \mathbb N^. $$ [...] – Akira Nov 20 '22 at 10:14
  • [...] Then $|K_n|{L{p}(\mu, X)^} \le |H|{L{p}(\mu, X)^}$ and $K_n \to H$ pointwise. It follows that $|\varphi(K_n)|{L{q}(\mu, X^)} \le |\varphi(H)|{L{q}(\mu, X^)}$. However, I could not get how $\varphi(K_n) \to \varphi(H)$ pointwise. Could you elaborate more? – Akira Nov 20 '22 at 10:14

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