Proof of Jordan Decomposition of a Bounded Linear Functional on the Continuous Functions with Compact Support

Throughout, the functions $0$, $f+g$, $\min(g,f)$, $f-g$, $cf$, $f^{+}$ and $f^{-}$ formed below from elements of $C_{c}(X)$ lie in $C_{c}(X)$ by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (i) and Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (ii). Write $M=\|\Lambda\|_{*}$; the dual norm is a bound for $\Lambda$,

$$|\Lambda g|\le M\,\|g\|_{u}\qquad(g\in C_{c}(X)).$$

Let $C_{c}^{+}$ be the set of $f\in C_{c}(X)$ with $f\ge0$.

The supremum. Let $f\in C_{c}^{+}$ and let $S(f)=\{\Lambda g:g\in C_{c}(X),\ 0\le g\le f\}$. It contains $\Lambda0=0$. If $0\le g\le f$, then $|g(x)|\le f(x)\le\|f\|_{u}$ for every $x\in X$, so $\|g\|_{u}\le\|f\|_{u}$, and $\Lambda g\le M\|g\|_{u}\le M\|f\|_{u}$. Thus $S(f)$ is nonempty and bounded above, and has a supremum $p(f)=\sup S(f)$. Since $0\in S(f)$ and $\Lambda f\in S(f)$,

$$0\le p(f)\le M\,\|f\|_{u},\qquad\Lambda f\le p(f)\qquad(f\in C_{c}^{+}).\tag{1}$$

Moreover $S(0)=\{\Lambda0\}$, so $p(0)=0$.

Additivity on nonnegative functions. Let $f_{1},f_{2}\in C_{c}^{+}$. If $0\le g_{1}\le f_{1}$ and $0\le g_{2}\le f_{2}$, then $0\le g_{1}+g_{2}\le f_{1}+f_{2}$, so $\Lambda g_{1}+\Lambda g_{2}=\Lambda(g_{1}+g_{2})\le p(f_{1}+f_{2})$; taking suprema over $g_{1}$ and then over $g_{2}$ gives $p(f_{1})+p(f_{2})\le p(f_{1}+f_{2})$. Conversely, let $0\le g\le f_{1}+f_{2}$, and put

$$g_{1}=\min(g,f_{1}),\qquad g_{2}=g-g_{1}.$$

Then $0\le g_{1}\le f_{1}$ and $g_{2}\ge0$. At a point $x$ with $g(x)\le f_{1}(x)$ we have $g_{2}(x)=0\le f_{2}(x)$, and at any other point $g_{2}(x)=g(x)-f_{1}(x)\le f_{2}(x)$; so $0\le g_{2}\le f_{2}$. Hence $\Lambda g=\Lambda g_{1}+\Lambda g_{2}\le p(f_{1})+p(f_{2})$, and taking the supremum over $g$,

$$p(f_{1}+f_{2})=p(f_{1})+p(f_{2})\qquad(f_{1},f_{2}\in C_{c}^{+}).\tag{2}$$

Homogeneity on nonnegative functions. Let $f\in C_{c}^{+}$ and $c\ge0$. For $c=0$, $p(0f)=p(0)=0$. For $c>0$, $g\mapsto cg$ is a bijection from $\{g\in C_{c}(X):0\le g\le f\}$ onto $\{g\in C_{c}(X):0\le g\le cf\}$, and $\Lambda(cg)=c\Lambda g$; hence $S(cf)=\{cs:s\in S(f)\}$ and

$$p(cf)=c\,p(f)\qquad(f\in C_{c}^{+},\ c\ge0).\tag{3}$$

Extension to $C_{c}(X)$. Every $f\in C_{c}(X)$ is $f=f^{+}-f^{-}$ with $f^{+},f^{-}\in C_{c}^{+}$. If also $f=u-v$ with $u,v\in C_{c}^{+}$, then $f^{+}+v=u+f^{-}$, and (2) gives $p(f^{+})+p(v)=p(u)+p(f^{-})$, that is,

$$p(f^{+})-p(f^{-})=p(u)-p(v).\tag{4}$$

Define $\Lambda^{+}f=p(f^{+})-p(f^{-})$ for $f\in C_{c}(X)$. For $f\in C_{c}^{+}$ we have $f^{+}=f$ and $f^{-}=0$, so $\Lambda^{+}f=p(f)$, which is the displayed supremum formula.

Linearity and positivity of $\Lambda^{+}$. Let $f,g\in C_{c}(X)$. Since $f+g=(f^{+}+g^{+})-(f^{-}+g^{-})$ with both terms in $C_{c}^{+}$, (4) and (2) give

$$\Lambda^{+}(f+g)=p(f^{+}+g^{+})-p(f^{-}+g^{-})=\Lambda^{+}f+\Lambda^{+}g .$$

Let $c\in\mathbb{R}$. If $c\ge0$, then $cf=cf^{+}-cf^{-}$, and (4) and (3) give $\Lambda^{+}(cf)=c\,p(f^{+})-c\,p(f^{-})=c\Lambda^{+}f$. If $c<0$, then $cf=(-c)f^{-}-(-c)f^{+}$, and likewise $\Lambda^{+}(cf)=(-c)p(f^{-})-(-c)p(f^{+})=c\Lambda^{+}f$. So $\Lambda^{+}$ is a linear functional on $C_{c}(X)$, and it is positive by (1).

The negative part. Define $\Lambda^{-}f=\Lambda^{+}f-\Lambda f$ for $f\in C_{c}(X)$. As a difference of linear functionals, $\Lambda^{-}$ is a linear functional, and $\Lambda f=\Lambda^{+}f-\Lambda^{-}f$ for every $f\in C_{c}(X)$. For $f\in C_{c}^{+}$, (1) gives $\Lambda^{-}f=p(f)-\Lambda f\ge0$, so $\Lambda^{-}$ is positive. For $f\in C_{c}^{+}$ and $0\le g\le f$ we have $0\le f-g\le f$, hence $\|f-g\|_{u}\le\|f\|_{u}$ as in the first step, and

$$\Lambda g-\Lambda f=-\Lambda(f-g)\le M\,\|f\|_{u};$$

taking the supremum over $g$ gives $\Lambda^{-}f=p(f)-\Lambda f\le M\|f\|_{u}$.

Bounds. Let $Q$ be $\Lambda^{+}$ or $\Lambda^{-}$. By (1) and the previous step, $0\le Qf\le M\|f\|_{u}$ for every $f\in C_{c}^{+}$. Let $f\in C_{c}(X)$. For every $x\in X$, $0\le f^{\pm}(x)\le|f(x)|\le\|f\|_{u}$, so $\|f^{\pm}\|_{u}\le\|f\|_{u}$. By linearity $Qf=Qf^{+}-Qf^{-}$, a difference of two numbers in $[0,M\|f\|_{u}]$, so

$$|Qf|\le M\,\|f\|_{u}\qquad(f\in C_{c}(X)).$$

Thus $M\ge0$ is a bound for $Q$: $Q$ is bounded, so $Q\in C_{c}(X)^{*}$, and $M$ lies in the set of nonnegative bounds of $Q$, whose infimum is $\|Q\|_{*}$ by The Dual Norm (i). Hence $\|\Lambda^{\pm}\|_{*}\le M=\|\Lambda\|_{*}$, which completes the proof of (i).