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Suppose the following situation. We have linear functional $\ell$ on the space $H(\mathbb{C}^n)$ of entire function and wish to find a representation for $l$ with integration against a complex Borel measure \mu with compact support. i.e:

$$\ell(f)=\int f d\mu\quad \forall f\in H(\mathbb{C}^n).$$

Well, my text suggests to use Hahn-Banach theorem and then Riesz representation. But I dont rly know how. Should $l$ be extended to $C(\mathbb{C}^n)$? And which version of Riesz representation theorem is to use?

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Yes, you extend $l$ to $\lambda \in C(\mathbb{C}^n)^\ast$. Then, since $\lambda$ is continuous, there is a compact $K \subset \mathbb{C}^n$ with

$$\lvert \lambda(f)\rvert \leqslant M\cdot \max \left\{\lvert f(z)\rvert : z \in K\right\},$$

and you use the Riesz representation theorem for continuous linear functionals on $C(K)$ (by Tietze's extension theorem, the restriction $\rho \colon C(\mathbb{C}^n) \to C(K)$ is surjective, and the kernel of $\rho$ is contained in $\ker\lambda$, so the induced functional on $C(K)$ is well-defined and uniquely determined by $\lambda$), which gives you a representing Borel measure $\mu$ on $K$.

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    $\begingroup$ Yes, precisely (should have mentioned Tietze's theorem too, sorry). $\endgroup$ Commented Mar 29, 2014 at 17:57
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    $\begingroup$ It is such a space. For compact $K$, you have $C(K) = C_0(K)$, since given any $f\in C(K)$ and $\varepsilon > 0$, there is a compact set $M\subset K$ (namely $K$ itself) such that $\lvert f(z)\rvert < \varepsilon$ for all $z\notin M$ (vacuously, here). We could however also use the $C_0(U)$ version for an open subset of $\mathbb{C}^n$ by multiplying with a cut-off function $\varphi$ and taking $U = \{ z : \varphi(z) > 0\}$. Then one needs to see that $\{\varphi\cdot f : f \in C(\mathbb{C}^n)\}$ is dense in $C_0(U)$, which isn't too hard either. $\endgroup$ Commented Mar 29, 2014 at 18:16
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    $\begingroup$ $\lambda$ (and a fortiori $l$) does only depend on the values on $K$. If $f$ and $g$ coincide on $K$, then $\lambda(f) = \lambda(g)$, since $\sup \{\lvert f(z)-g(z)\rvert : z\in K\} = 0$. The measure you get does indeed represent $l$, you have $$l(f) = \int_K f\,d\mu$$ for all $f\in H(\mathbb{C}^n)$ by construction. $\endgroup$ Commented Mar 29, 2014 at 18:58
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    $\begingroup$ You extend the real and imaginary part of the function. A continuous complex-valued function is just a pair of continuous real-valued functions. $\endgroup$ Commented Mar 30, 2014 at 0:46
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    $\begingroup$ No problem, @cQQkie, while I'm here, I'm here and notifications don't bother. When I'm not here, I'm away and notifications don't bother ;) $\endgroup$ Commented Mar 30, 2014 at 0:50

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