Let $X$ be any normed vector lattice.

How to show that $\| x \| = \| |x| \|$ holds for all $x \in X$?

I thought its proof is trival. Because from the property of a lattice norm $\| \cdot \|$, we have $\|x\| \leq \|y\|$ whenever $|x| \leq |y|$ in $X$. So, we could set $y := |x|$ and obviously get $ |x| \leq |y|$ (precisely, $|x| = |y|$). Then, using the property of a lattice norm gives us $\| x \| = \| y \| = \| |x| \|$.

Is it a correct proof of it? Thanks in advance!

  • $\begingroup$ So, taking $|x|= (x\lor 0)\, +\, (-x\lor 0)$ in the vector lattice, I don't see what guarantees that $|x|\le |y|$ implies $\|x\|\le\|y\|$. Also, can you give us a quick definition of reference to normed vector lattices? $\endgroup$ – Berci Mar 21 '17 at 23:38
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    $\begingroup$ "Then, using the property of a lattice norm gives us..." Using what property? It doesn't seem that trivial to me... you have $|x|=x\vee(-x)$ or worded a different way $|x|=x^++x^-$. As for the norm you are using, it isn't clear which norm it is, but I would imagine it would be $\|x\|_e=\inf\{\lambda\geq 0:|x|\leq \lambda e\}$. From here it is helpful to note that $|(|x|)|=|x\vee(-x)|=(x\vee(-x))\vee(-(x\vee(-x))$ Now, can you simplify the right? $\endgroup$ – JMoravitz Mar 21 '17 at 23:38
  • $\begingroup$ @Berci the definitions in the book I'm reading suggest that a normed lattice is a vector lattice with the property that $|x|\leq |y|$ implies $\|x\|\leq \|y\|$, so it should be guaranteed by definition. $\endgroup$ – JMoravitz Mar 21 '17 at 23:42
  • $\begingroup$ @JMoravitz we don't need to specify the lattice norm on $X$. In fact, for any $x \in X$, we have $x = x^{+} - x^{-} \leq |x| = x^{+} + x^{-}$. Also, $| x^{+} - x^{-} | = | x^{+} + x^{-} |$ holds. By the monotonic property of the lattice norm, $| x^{+} - x^{-} | = | x^{+} + x^{-} |$ implies that $\| x^{+} - x^{-} \| = \| x^{+} + x^{-} \|$. $\endgroup$ – Paradiesvogel Mar 21 '17 at 23:54
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    $\begingroup$ Your proof is correct. Let me write it down slightly more detailed so that every step is clear: You are basically using the fact that for every $x\in X$, $|x|=|\ |x|\ |$. Then the inequality $|x|\leq |\ |x|\ |$, combined with the lattice property, gives that $\|x\|\leq \|\ |x|\ \|$. Similarly, $|\ |x|\ | \leq |x|$ gives the opposite inequality. $\endgroup$ – tree detective Mar 22 '17 at 13:56

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