In this post of Prof Tao, it shows a lemma about the decomposition of bounded linear functional and a related exercise. The content is

Lemma 9 (Jordan decomposition for functions) Let $I \in C_c(X \rightarrow \bf{R})^*$ be a (real) continuous linear functional. Then there exist positive linear functions $I^+, I^- \in C_c(X \rightarrow \bf{R})^*$ such that $I = I^+ - I^-$.

Exercise 15 Show that among all possible choices for the functionals $I^+, I^-$ appearing in the above lemma, there is a unique choice which is minimal in the sense that for any other functionals $\tilde I^+, \tilde I^-$ obeying the conclusions of the lemma, one has $\tilde I^+(f) \geq I^+(f)$ and $\tilde I^-(f) \geq I^-(f)$ for all $f \in C_c(X \rightarrow \bf{R})$.

I have a question about the existence of the so-called minimal decomposition, because roughly speaking I think THE sense of ordering between two decompositions defined in the exercise is not necessarily applicable to any two decompositions.

More precisely, given any two decompositions $I=I^+_1-I^-_1=I^+_2-I^-_2$, the only thing we can get is that the equality $I^+_1(f)-I^+_2(f)=I^-_1(f)-I^-_2(f)$ holds for any $f\in C_c(X)$. But this fact does not imply that $I^+_1(f)-I^+_2(f)\ge 0$($\le 0$) for any $f\in C_c(X)$. It can take positive or negative values for different $f$. It is required only that $I^+_1-I^+_2$ and $I^-_1-I^-_2$ are the same general linear functional. Then we can't say which decomposition is greater or less than which, right ?

In my opinion, this ordering is a partial order among all possible decompositions. Right ?

(I know the uniqueness part is easy to prove if two minimal decompositions exist, because the difference of two minimal positive functionals must be $0$. Then they must be equal.)


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