Let $X$ be a locally compact Hausdorff space. If you want, assume $X$ is also a metric space or even $X=\mathbb{R}^n$. Let $L^\infty(X)$ be the set of all bounded, Borel measurable, complex valued functions on $X$ equipped with the $sup$ norm. Suppose I have a bounded linear functional $\lambda: L^\infty(X) \to \mathbb{C}$. Now I read about the Riesz representation theorem that says that complex measures on $X$ are the same as bounded linear functional $ C_c(X) \to \mathbb{C}$. (Here $C_c(X)$ are continuous compactly supported complex valued functions, and we could replace this with continuous complex valued functions "vanishing at infinity" since the former is dense in the latter.). Therefore $\lambda$ determines a complex measure. This complex measure has the form $h \cdot d\mu$ for $h$ a Borel measurable function taking values on the unit circle in $\mathbb{C}$ and $\mu$ a finite positive Borel measure. Thus it is easy to see that this complex measure, in turn, gives a bounded linear function $\lambda'$ on $L^\infty(X)$.
Now here's a stupid question: Is it true that $\lambda=\lambda'$? If so, how to see that? If not, are there some kind of criteria that tell you that $\lambda$ will be "nice" and line up with the complex measure it determines?
Edit: Ok it seems from this post that the answer is no in general: Bounded linear functionals on $L^\infty$.. An in hindsight obvious application of Hahn Banach.
I will wait to see if there is some interesting answer to my second question and maybe delete if not.
Best Answer
No. Consider the case that $X=\mathbb{R}$. Let $E=\{f\in L^{\infty}(X)\mid\lim_{x\rightarrow+\infty}f(x)\mbox{ exists}\},$ then $E$ is a vector subspace of $L^{\infty}(X)$. Moreover, $C_{0}(X)\subset E\subset L^{\infty}(X)$. (Here, $A\subset B$ means that $A$ is a proper subset of $B$ and $A\subseteq B$ means that $A$ is a subset of $B$).
Define $\theta:E\rightarrow\mathbb{C}$ by $\theta(f)=\lim_{x\rightarrow+\infty}f(x)$. Clearly $\theta$ is linear and $|\theta(f)|\leq||f||$. That is, $\theta$ is a bounded linear functional on $E$ with $||\theta||\leq1$. Note that by considering $f\equiv1$, we conclude that $||\theta||=1$.
By Hahn-Banach theorem, there exists a bounded linear functional $\lambda:L^{\infty}(X)\rightarrow\mathbb{C}$ such that $\lambda|_{E}=\theta$ and $||\lambda||=||\theta||=1$. However, $\lambda|_{C_{0}(X)}=0$. For if $f\in C_{0}(X)$, then $\lambda(f)=\theta(f)=\lim_{x\rightarrow\infty}f(x)=0$. Hence, the regular complex Borel measure $\mu$ induced by $\lambda|_{C_{0}(X)}$ via Riesz-representation theorem is $\mu=0$. Define $\phi:L^{\infty}(X)\rightarrow\mathbb{C}$ by $\phi(f)=\int fd\mu$, then $\phi$ is a bounded linear functional. However, $\phi=0$ because $\mu=0$. This shows that $\lambda\neq\phi$.