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Let $\Omega \subset \mathbb{R}^n$ be open and bounded with smooth boundary. For $f \in L^1(\Omega)$, define $$ \|D_1 f\|_M(\Omega) =\inf\left\{\liminf_{k\to\infty}\int_\Omega |\nabla f_k|\,dx \mid f_k \to f \text{ in } L^1(\Omega),\ f_k \in \text{ Lip }(\Omega)\right\}. $$ Here $\text{Lip}(\Omega)$ is the set of Lipschitz functions on $\Omega$. Note that by Rademacher's Theorem, for $f \in \text{Lip}(\Omega)$, $\nabla f$ exists Lebesgue-a.e. My question is, is $\|D_1 f\|_M(\Omega)$ the same as $\int_\Omega |Df|$ in general? I have a feeling the answer is ''no'', because if it is ''yes'', people would probably use this as the definition of bounded variation instead of the usual definition, which I find more complicated.

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Is $\omega$ supposed to be a test function? – Jose27 May 28 '12 at 2:51
Oops, the $\omega$ doesn't belong there. I removed it. Thanks. – Stefan Smith May 28 '12 at 7:12
up vote 2 down vote accepted

Yes, this is another way to introduce the BV norm, sometimes called Miranda's definition. People do use it, but it does not mean the distributional definition can be forgotten. It's not a bad thing to have two or more definition of the same class. For example, they might generalize in different ways when we move beyond Euclidean spaces. This dissertation is relevant.

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Thanks! I think this is a record for the longest it's taken me to get a useful answer! I have done a fair amount of reading about BV functions, and I'm surprised this definition is not more popular. I kind of discovered it myself (without knowing for sure if it is equivalent to the usual definition). I don't think Giusti's book even mentions it. In my opinion, it is much simpler than the standard definition. – Stefan Smith Mar 7 '13 at 22:50

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