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!


  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*}

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.