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I'm struggling with following problem:

Let $\mathcal{F}$ be a vector lattice of bounded functions on a set $X$ such that $1\in\mathcal{F}$. Suppose that we are given a linear functional $L$ on $\mathcal{F}$ that is continuous with respect to $\lvert|f|\rvert=\sup_\Omega|f(x)|$. Then $L$ can be represented in the form $L=L^+-L^-$, where $L^+\geq0$, $L^-\geq0$ and for all nonnegative $f\in\mathcal{F}$ one has $$ L^+(f)=\sup_{0\leq g\leq f}L(g)\quad\text{ und }\quad L^-(f)=-\inf_{0\leq g\leq f}L(g)$$

Let $L^+$ be given in the stated way. I have already proven:

  1. $|L^+(f)|<\infty$
  2. $\forall\,f\geq0,\forall t\geq0\,:\,L^+(tf)=tL^+(f)$
  3. $\forall\,f,g\in\mathcal{F}\,:\,f,g\geq0\,:\,L^+(f+g)=L^+(f)+L^+(g)$

Now, to extend the functional $L^+$ to all $f\in\mathcal{F}$, one defines $$L^+_*(f):=L^+(f^+)-L^+(f^-),$$

where as usual $f^+=\max(f,0)\geq0$ and $f^-=-\min(f,0)\geq0$. For this extended functional I have proven $L^+_*(f+g)=L^+_*(f)+L^+_*(g)$ for all $f$, $g\in\mathcal{F}$.

Now, I have 2 questions:

  1. How can I show that $L^+_*(tf)=tL^+_*(f)$ for all $f\in\mathcal{F}$ and all real numbers $t$? In my opinion, $L^+(tf^\pm)$ is not even defined for $t<0$.
  2. If I define $L^-:=L^+-L$, then $L^-$ is obviously positive. How can I show the stated formula $L^-(f)=\displaystyle-\inf_{0\leq g\leq f}L(g)$?

Any help would be highly appreciated, thank you very much in advance for your efforts!

Florian

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  1. You've already proven the case for $t \ge 0$, so it remains to consider the case $t < 0$. In fact, it suffices to consider only the case $t = -1$. In order to do so, you need to be able to simplify $(-f)^+$ and $(-f)^-$. As it turns out, $(-f)^+ = f^-$ and $(-f)^- = f^+$, by some basic vector lattice operations (or by the specific definition of the positive and negative parts in the case of $\mathcal{F}$). Therefore, \begin{align*} L^+_*(-f) &= L^+_*((-f)^+) - L^+_*((-f)^-) \\ &= L^+_*(f^-) - L^+_*(f^+) \\ &= -L^+_*(f). \end{align*}

  2. If you define $L^-$ in such a way, and fix $f \ge 0$, then \begin{align*} L^-(f) &= L^+(f) - L(f) \\ &= \sup\limits_{0 \le g \le f} L(g) - L(f) \\ &= \sup\limits_{0 \le g \le f} L(g - f) \\ &= -\inf\limits_{0 \le g \le f} L(f - g) \\ &= -\inf\limits_{0 \le f - h \le f} L(h) ~ ~ ~ \ldots \text{making substiution } h = f - g \\ &= -\inf\limits_{0 \le h \le f} L(h). \end{align*}

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    $\begingroup$ You're welcome! Thank you for the sorely needed reputation. $\endgroup$ – Theo Bendit Jun 16 '15 at 11:35
  • $\begingroup$ Dear Mr. Bendit, could you help me once more? If I define $|L|:=L^++L^-$, how can I show $$|L|(f)=\sup_{0\leq|g|\leq f}|L(g)|?$$ No idea why this is so hard for me... $\endgroup$ – Florian Jun 17 '15 at 13:22
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    $\begingroup$ First rewrite the right hand side as $\sup\limits_{-f \le g \le f} L(g)$. Then, rewrite the left hand side as $\sup\limits_{0 \le p \le f} L(p) + \sup\limits_{0 \ge q \ge -f} L(q) = \sup\limits_{-f \le q \le 0 \le p \le f} L(p + q)$. With these limits, $p + q$ lies between $-f$ and $f$, so left side is less than right side. On the other hand, any $g$ between $-f$ and $f$ satisfies $-f \le -g^- \le 0 \le g^+ \le f$, and $L(g) = L(g^+ + (-g^-))$, so we have equality. If you want a better-formatted answer, you may have to actually ask a new question. $\endgroup$ – Theo Bendit Jun 17 '15 at 14:56

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