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I try to found a definition of a function $$f(x)\colon\mathbb R^m\to \mathbb R^n$$ that use the norm. Is the formula below correct? $$TV=\sup\sum_{i=1}^k \|f(x_i)-f(x_{i-1})\|.$$ with k any finite integer

Elisa
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No, the definition above only works for $f: [a,b] \rightarrow \mathbb{R}^n$, in other words, where the domain is a real interval. Because what is the supremum over? On the real line, the supremum over partitions of the interval is the appropriate supremum, but over $\mathbb{R}^m$, the notion of "partition" doesn't quite work (or, if it can be generalized in a way I'm unaware of, it must be rather complicated).

To properly generalize to vector arguments, one needs to instead characterize the variation norm by weak derivatives. See Wikipedia for details.

Christopher A. Wong
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