• iAvicenna@lemmy.world
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    1 day ago

    Look it is so simple, it just acts on an uncountably infinite dimensional vector space of differentiable functions.

    • gandalf_der_12te@discuss.tchncs.de
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      15 hours ago

      fun fact: the vector space of differentiable functions (at least on compact domains) is actually of countable dimension.

      still infinite though

      • iAvicenna@lemmy.world
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        12 hours ago

        Doesn’t BCT imply that infinite dimensional Banach spaces cannot have a countable basis