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Let $W^{1,1}_0(0,1)$ be the space of functions on the interval $(0,1)$ that vanish at the boundary with the standard $W^{1,1}(0,1)$-norm. $$||f|| = \int_0^1 |f(x)| \, dx + \int_0^1 |f'(x)| \, dx.$$

  1. Is this space a Banach lattice? It appears to satisfy all the conditions listed here. Two functions $f, g$ satisfy $f < g$ if $f(x) < g(x)$ for all $x \in (0,1).$

  2. What about the condition $$|| f || = || \, \, |f| \, \, ||$$? Is this obvious? Does one need to use smooth approximations?

3.If $f$ and $g$ are disjoint, then they satisfy $$ || f+g || = || f|| + ||g ||.$$ So is $W^{1,1}_0(0,1)$ an abstract $L^1$ space?

Any reference will be greatly appreciated.

Arctic Char
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doc
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  • The condition on the absolute values is not trivial, but well-known (see for instance Evans's book "Partial Differential Equations"). – Jose27 Jul 24 '21 at 22:47
  • Thank you. I tried looking at Evans. Can you point to a section or a theorem in Evans? – doc Jul 25 '21 at 14:44
  • This is typically proved as a corollary of chain rules for Sobolev functions. Evans only has exercise on it:Ex 17 in 5.10 – daw Jul 26 '21 at 06:27
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    Sometimes (e.g. https://en.wikipedia.org/wiki/Banach_lattice ) the condition ``$|f|\le |g|$ implies $|f|\le |g|$'' is required for Banach lattices, which is not true for $W^{1,1}$ – daw Jul 26 '21 at 06:29
  • Yes, I have noticed that this other condition is required. Are these two definitions equivalent. They do not appear to be. – doc Jul 26 '21 at 12:26

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