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$\textbf{Background:}$

When $X$ is a locally compact Hausdorff space, there is a Jordan decomposition theorem for the dual of $C_{0}(X,\mathbb{R})$. In particular, if $I\in C_{0}(X, \mathbb{R})^{*}$, then we may find positive linear functionals $I^{+}$ and $I^{-}$ so that $I = I^{+} - I^{-}$.

One might guess that something like this is always true. If $B$ is a Banach space consisting of real-valued functions $f:X\rightarrow\mathbb{R}$, then we may define positive functionals of $B^{*}$ to be the functionals $I\in B^{*}$ so that if $f\ge 0$ then $I(f)\ge 0$.

$\textbf{Question:}$

(1) With $B$ as above, can you always write $I\in B^{*}$ as $I^{+} - I^{-}$? Is there an illustrative counterexample?

(2) If this is false in general, is there an example of a more general theorem than the one given at the very beginning of this post?

$\textbf{Remark:}$ I believe that Decomposition of functionals on sobolev spaces might give a counterexample, but I have not studied the duality of Sobolev spaces.

user3281410
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1 Answers1

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Say $B=C^1([-1,1])$ and $If=f'(0)$.

It might be interesting to determine hypotheses on $B$ that imply every linear functional is the difference of two positive functionals.

David C. Ullrich
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  • I think I see what you meant. If $D$ is the differentiation functional and $D = P - N$ for some $P,N$ positive, then we would have to have $P(1) = N(1)$, and for $f\ge 0$, we would have $P(f - ||f||\infty)\le 0$, which implies $P(f)\le ||f||\infty P(1)$, which is a contradiction, since we can make the derivative of a function large without making the function large. – user3281410 Feb 26 '18 at 23:21
  • When you say it "might be interesting," do you mean that you think it is interesting and also know the answer or that you think it's interesting but do not know? My motivation for this question was looking at the space of bounded measurable functions with the uniform norm. The continuous functions on this space is the space of finitely additive measures, which has a Jordan (but not a Hahn) decomposition theorem, I think. – user3281410 Feb 26 '18 at 23:25
  • So my next guess will be that we might have a Jordan decomposition when $B$ is any space of real functions with the uniform norm. Also, thanks for your help. – user3281410 Feb 26 '18 at 23:49
  • @user3281410 I don't know. – David C. Ullrich Feb 27 '18 at 01:09