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The Riesz Representation Theorem: Let $X$ be a locally compact Hausdorff space and $I$ a positive linear functional on $C_c(X)$ (the set of complex functions with compact support). Then there is a unique positive measure ${\mu}$ on $\mathcal{B}(X)$, the Borel $\sigma$-algebra associated with the topology on $X$, for which $$ I(f)= \int_X f \,d{\mu} \text{ for all } f \in C_c(f) $$

Suppose that we can proof this theorem for real functionals , how can we extend it for complex functionals ?

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This can be done! It's a remarkable fact that the Reisz Representation theorem for complex linear functionals is substantially stronger than the one for real functionals. In the complex case, the theorem incorporates the structure of complex measures on $X$ as a normed linear space (by total variation norm). Additionally, since complex measures have bounded total variation, the $\sigma$-compact hypothesis in the real Riesz Representation theorem vanishes here.

In particular, the complex case of the Riesz Representation Theorem says that the map $$ \Phi: \{\text{regular complex Borel measures on } X\} \rightarrow C_0(X)^*$$ given by $$\Phi(\mu)(f) = \int_X fd\mu$$ is an isometric isomorphism. The extra bit of algebraic content of this result from the real case is immensely useful, and makes the Riesz Representation Theorem a fundamental tool in functional analysis.

Proving this version of the Riesz Representation Theorem is not just an easy extension of the real case of the result. For a reference, chapter 6 of Rudin's Real and Complex Analysis proves this. The proof given by Rudin uses the real case of the RRT and the fact that for a real, $\sigma$-finite measure $\lambda$, the dual of $L^1(X, \lambda)$ can be identified with $L^\infty(X, \lambda)$, which is a nontrivial result.

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