# Uniqueness in the Riesz representation theorem for the dual of $C(X)$ in the book by Royden

I'm reading the book Real Analysis 4e by Royden on the Riesz representation theorem for the dual of $C(X)$. I have two problems about the proof of this theorem. The book states the theorem this way in page 464,

Let $X$ be a compact Hausdorff space and $C(X)$ the linear space of real-valued functions on $X$, normed by maximum norm. Define the operator $T: \textrm{Radon}(X)\to[C(X)]^*$ by setting, for $v\in \textrm{Radon}(X)$, $$T_v(f)=\int_X fdv$$ for all $f\in C(X)$. Then T is a linear isometric isomorphism of $\textrm{Radon}(X)$ onto $[C(X)]^*$.

Here $\textrm{Radon}(X)$ is the collection of all signed Radon measures on $X$ equiped with the norm, $$\|\mu\|=\mu^+(X)+\mu^-(X)$$ where $\mu^+$ and $\mu^-$ are unique Randon measures according Jordan decomposition.

In its proof, 1) it begins by showing that any bounded linear functional $L$ in $[C(X)]^*$ can be represented as the difference of two positive linear functionals as $L=L_1-L_2$ with $\|L\|=L_1(1)+L_2(1)$, without proving the uniqueness of the $L_1$ and $L_2$. Then according the Riesz-Markov theorem which states that each positive linear functional $L_1$ can be uniquely expressed as $$L_1(f)=\int_X fd\mu_1$$ with a $\mu_1$ being a Radon measure, each bounded linear functional $L$ can be written as $$L(f)=L_1(f)-L_2(f)=\int_X fd\mu_1-\int_X fd\mu_2=\int_X fd(\mu_1-\mu_2)=\int_X fd\mu$$ with $\mu=\mu_1-\mu_2$ being a signed Randon measure. Thus $L=T_{\mu}$ and the operator $T$ is onto. Up to now, I can follow the argument. The following is what confuses me. It says We infer from this and Proposition 11 that the representation of $L$ as the difference of positive linear functionals is unque. The proposition 11 states that

Let X be locally compact Hausdorff space and $\mu_1,\mu_2$ be Radon measures on $X$ for which $$\int_X fd\mu_1=\int_X fd\mu_2$$ for all compact supported continuous function $f$ in $C_c(X)$. Then $\mu_1=\mu_2$.

I don't understand how the uniqueness of the decomposition of $L$ can follow from this proposition and the fact that $T$ is onto:-(.

2) The proof continues by showing that $$\|L\|=\mu_1(X)+\mu_2(X)=\|\mu\|$$ which I understand. It then immediately jumps to the conclusion that Therefore T is an isometric isomorphism, which I have a little problem about.

I tried to fill the missing details about the final jump myself. If I accept the argument before the final jump, then I think the proof, in summary, shows that any bounded linear functional $L$ can uniquely determine a signed Radon measure $\mu$ and thus the operator $T$ is one-to-one and onto. Also, obviously, $T$ is linear. Therefore $T$ has a linear inverse $T^{-1}$. The last formula $\|L\|=\|\mu\|$ leads to the fact that $T$ and $T^{-1}$ both have a finite/bounded norm according the definition of the norm of operators ($\|T\|=sup_{\mu}\frac{\|T_{\mu}\|}{\|\mu\|}$ in this theorem). Thus, both $T$ and $T^{-1}$ are bounded linear operators, i.e., continuous linear operators. Thus $T$ is an isomorphism. $\|T\|$ happens to be $1$. This means it is also isometric isomorhphism which preserves the norms of $L$ and $\mu$. Is my understanding about the final jump right ?

• What exactly do you call "the final jump"? I don't have the book at hand, but if you didn't misunderstand and misrepresent the argument, it's not quite correct. – Daniel Fischer Oct 29 '16 at 18:23
• @DanielFischer By "final jump", I mean the final conclusion that T is a isometric isomorhpism in the second to last paragraph. In the book, it concludes that T is a isometric isomorphism immediately after showing $\|T\|=\|\mu\|$. – Hua Oct 30 '16 at 2:25
• Well, $\lVert T\mu\rVert = \lVert\mu\rVert$ for all $\mu$ is precisely the definition of $T$ being isometric (for linear $T$). This already proves that $T$ is injective, and that $T$ is surjective was shown before. There are some issues in the proof, at least in your representation of it, however. It is not true that the representation of $L$ as the difference of two positive functionals is unique (at least for nonempty $X$). Take any positive functional $P\neq 0$, then $L = L_1 - L_2 = (L_1 + P) - (L_2 + P)$ are two different representations of $L$ as a difference of positive functionals. – Daniel Fischer Oct 30 '16 at 13:13
• @DanielFischer Thanks for your tips and I may find what is the issue. In its proof of showing the decomposition of bounded linear functional, it also gives a property that $\|L\|=L_1(1)+L_2(1)$. So in this case, your example will increase the norm. This may be the issue. But I still don't know why the proposition 11 can guarantee the uniqueness. I'll check this in detail. – Hua Oct 30 '16 at 16:23
• The point is that there is a "minimal" decomposition of $L$ into a difference of positive functionals. Let's call a decomposition $L = L^+ - L^-$ of $L$ into a difference of positive functionals minimal if for every decomposition $L = L_1 - L_2$ into positive functionals the two functionals $L_1 - L^+$ and $L_2 - L^-$ are positive. Then it's clear that if a minimal decomposition exists, this is uniquely determined (since $T$ and $-T$ both positive implies $T=0$). Once you have any decomposition $L=L_1 - L_2$, you get a minimal decomposition from the Hahn decomposition of $\mu = \mu_1 - \mu_2$. – Daniel Fischer Oct 30 '16 at 17:39

## 1 Answer

I've lost track a little of what is in question here. I think that if we show the minimality of the constructed decomposition of $L$ as the difference of two positive linear functionals, the rest of the proof will be clear. If not, please request further elaboration.

Since I don't have access to Royden's book at the moment, I don't know how he constructs the decomposition. I expect it will be the standard method, though, which I first will sketch:

Let $P = \{ f \in C(X) : f \geqslant 0\}$. Then we start by constructing a functional $\Lambda_0 \colon P \to [0,+\infty)$ that dominates $L$. We define

$$\Lambda_0(f) = \sup \{ L(g) : g \in C(X), \, \lvert g\rvert \leqslant f\}.$$

Since $\lvert \pm f\rvert \leqslant f$ for $f\in P$, we have $\Lambda_0(f) \geqslant \max \{ L(f), L(-f)\} = \lvert L(f)\rvert \geqslant 0$, and since $\lvert L(g)\rvert \leqslant \lVert L\rVert\cdot\lVert g\rVert_\infty \leqslant \lVert L\rVert\cdot \lVert f\rVert_\infty$ for $\lvert g\rvert \leqslant f$, we have

$$0 \leqslant \Lambda_0(f) \leqslant \lVert L\rVert\cdot \lVert f\rVert_\infty$$

for $f\in P$.

Next one shows that $\Lambda_0(t\cdot f) = t\cdot \Lambda_0(f)$ for $t \in [0,+\infty)$ and $\Lambda_0(f + g) = \Lambda_0(f) + \Lambda_0(g)$ for $f,g \in P$. Thus one can extend $\Lambda_0$ to a positive linear functional $\Lambda_1$ on $C(X)$ by defining

$$\Lambda_1(f) = \Lambda_0(f^+) - \Lambda_0(f^-)$$

for $f \in C(X)$. One verifies that $\lVert\Lambda_1\rVert = \Lambda_1(1) = \lVert L\rVert$, and that $L_1 := \frac{1}{2}(\Lambda_1 + L)$ and $L_2 := \frac{1}{2}(\Lambda_1 - L)$ are positive linear functionals. Then one has the decomposition $L = L_1 - L_2$, and $\lVert L_1\rVert + \lVert L_2\rVert = L_1(1) + L_2(1) = \Lambda_1(1) = \lVert L\rVert$.

It remains to show the minimality of this decomposition. Suppose one has positive linear functionals $M_1, M_2$ with $L = M_1 - M_2$. We want to show that $M_1 - L_1 = M_2 - L_2$ are positive. So let $f \in P$, and $g \in C(X)$ with $\lvert g\rvert \leqslant f$. Then

\begin{align} L(g) &= M_1(g) - M_2(g) \\ &= M_1(g^+ - g^-) - M_2(g^+ - g^-) \\ &= M_1(g^+) - M_1(g^-) - M_2(g^+) + M_2(g^-) \\ &\leqslant M_1(g^+) + M_1(g^-) + M_2(g^+) + M_2(g^-) \\ &= M_1(\lvert g\rvert) + M_2(\lvert g\rvert) \\ &\leqslant M_1(f) + M_2(f), \end{align}

and thus $\Lambda_1(f) = \Lambda_0(f) \leqslant M_1(f) + M_2(f)$, from which we obtain

$$2L_1(f) = \Lambda_1(f) + L(f) \leqslant M_1(f) + M_2(f) + L(f) = M_1(f) + M_2(f) + M_1(f) - M_2(f) = 2M_1(f),$$

which is equivalent to the desired $M_1 - L_1 \geqslant 0$.

And by the minimality, we have

\begin{align} \lVert M_1\rVert + \lVert M_2\rVert &= M_1(1) + M_2(1)\\ &= L_1(1) + L_2(1) + \bigl((M_1-L_1)(1) + (M_2 - L_2)(1)\bigr)\\ &\geqslant L_1(1) + L_2(1) = \lVert L\rVert, \end{align}

with equality if and only if $0 = (M_1-L_1)(1) + (M_2-L_2)(1) = \lVert M_1 - L_1\rVert + \lVert M_2 - L_2\rVert$, i.e. $M_1 = L_1$ and therefore also $M_2 = L_2$.

While we do not have uniqueness of the decomposition of $L$ as the difference of two positive functionals (except in the trivial case $C(X) = \{0\}$, which happens for $X = \varnothing$) without side conditions, we have a unique decomposition with the additional condition that $\lVert L_1\rVert + \lVert L_2\rVert = \lVert L\rVert$.

The existence of this decomposition together with the Riesz-Markov theorem gives the existence of a Radon measure $\mu = \mu_1 - \mu_2$ with $T_\mu = T_{\mu_1} - T_{\mu_2} = L_1 - L_2 = L$ such that $\lVert\mu\rVert \leqslant \lVert\mu_1\rVert + \lVert\mu_2\rVert = \lVert L_1\rVert + \lVert L_2\rVert = \lVert L\rVert$. Since on the other hand one clearly has $\lVert T_\nu\rVert \leqslant \lVert\nu\rVert$ for every Radon measure $\nu$, it follows that $\lVert\mu\rVert = \lVert L\rVert$.

• Really thanks for such a detailed answer ! I read through quickly and I guess I got your idea. But I need some time to compare this with my previous confusion :-( I am preparing for an exam recently, so I may be absent time to time :-( By the way, there is typo at the end of paragraph starting with "Next one shows that", where $\|F\|$ should be $\|L\|$. I can't edit with less than 6 characters. – Hua Nov 13 '16 at 16:26
• Oh, I guess you there is another kind of misleading typo in the paragraph starting with "And by the minimality", where $L-1$ should be $L_1$. And I have a question about how you get the equality about the $2\|M_1-L_1\|$ in the same paragraph ? – Hua Nov 13 '16 at 16:45
• Thanks for spotting the typos. We get the $2\lVert M_1 - L_1\rVert$ from $M_2 - L_2 = M_1 - L_1$. so $(M_1 - L_1)(1) + (M_2 - L_2)(1) = \lVert M_1 - L_1\rVert + \lVert M_2 - L_2\rVert = 2\lVert M_1 - L_1\rVert$. – Daniel Fischer Nov 13 '16 at 17:26