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In a paper by Ginibre "Le problème de Cauchy pour des EDP semi-linéaires périodiques en variables d’espace" he utilizes phrases like

"des estimations linéaires assurant que l’application $\phi \rightarrow U(\cdot) \phi$ est bornée de $\mathcal{H}$ dans $X$"

(here $\mathcal{H}$ is Hilbert and $X$ is Banach) a lot. Can someone tell me what this exactly means? More precisely I'm having trouble with understanding "bornée de $\mathcal{H}$ dans $X$", so literally "bounded from $\mathcal{H}$ in $X$". Maybe it means that $\mathcal{H} \subset X$? Or something completely different?

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    $\begingroup$ It certainly means 2 things grouped in a too compact way : it is a mapping from $H$ to $X$ and its values are bounded, (i.e., $\phi(H)$ is bounded) $\endgroup$
    – Jean Marie
    Nov 11, 2019 at 12:36
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    $\begingroup$ It means this. $\endgroup$ Nov 11, 2019 at 12:39
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    $\begingroup$ @conditionalMethod Thanks, I know what a bounded operator is, that was unfortunately not the question :) $\endgroup$ Nov 11, 2019 at 13:07
  • $\begingroup$ @JeanMarie Ok that also makes sense in the context. So to be completely sure: $\phi \rightarrow U(\cdot) \phi$ is a mapping from $H$ to $X$ and bounded as a mapping? Follow-up question: What do you mean by $\phi(H)$ being bounded in $X$? Do you mean that $\sup_{x\in \phi(H)} \| x \|_X < \infty$? $\endgroup$ Nov 11, 2019 at 13:11
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    $\begingroup$ Take a look at the inequalities (2.1), (2.2), (3.4), (3.5), (3.10), (3.11), (3.12), (3.20), (3.21), ... $\endgroup$ Nov 11, 2019 at 13:16

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I'm not much of a French speaker, but I'd read this as "the map $\phi \mapsto U(\cdot)\phi$ is bounded from $\mathcal{H}$ to $X$." In other words, considering the context, the linear operator $\phi \mapsto U(\cdot) \phi$, which maps $\mathcal{H}$ into $X$, is a bounded operator with respect to the norms on $\mathcal{H}$ and $X$; we have $\|U(\cdot) \phi\|_X \le C \|\phi\|_{\mathcal{H}}$ for some constant $C$.

Note that it seems $\mathcal{H}$ is understood to be some space of functions on some set such as $\mathbb{R}^n$, and $U(\cdot)$ is a one-parameter family of linear operators, parametrized by some interval $I$, so $U(\cdot) \phi$ means the function $I \times \mathbb{R}^n \to \mathbb{R}$ defined by $(t,x) \mapsto (U(t)\phi)(x)$. Thus $X$ should be some space of functions on $I \times \mathbb{R}^n$.

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  • $\begingroup$ Yes, X is defined in that way. Thanks for clearing the confusion up a bit! $\endgroup$ Nov 11, 2019 at 13:42

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