Talk:Nemytskii operator
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In the section of the Boundless Theorem it may be added, that there exists a more general form where one concludes from , , that the Nemytskii operator is a bounded and continuous map from to .
This may belong to another article; but I have not found a better place yet.
Redundant Sections
[edit]This article, in its current form, gives three very similar definitions of the operator in question in its first three sections, which differ from each other only in slight details - I don't feel I am familiar enough with functional analysis to judge which one should be preferred, but surely the article should only give a single generalized definition and derive everything else from there? Yoneda-Emma (talk) 16:14, 8 January 2026 (UTC)