Abstract
A necessary and sufficient condition is given to ensure the boundedness of Fourier–Haar multiplier operators from $$L^1 ([0, 1], X)$$ to $$L^1 ([0, 1], Y)$$ , where X is an arbitrary finite dimensional Banach space and Y is an arbitrary Banach space. The Fourier–Haar multiplier sequences come not from $${\mathbb {R}}$$ , as in the classical case, but from the space of bounded operators from the Banach space X to the Banach space Y. Moreover, it is shown that this condition characterises the finite dimensionality of the Banach space X.
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