Let $f:\mathbb R\to\mathbb C$ be bounded and Borel measurable, $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$ and $$M_fg:=fg\;\;\;\text{for }g\in\mathcal D(M_f):=\left\{g\in L^2(\lambda;\mathbb C):\int|fg|\:{\rm d}\lambda<\infty\right\}.$$ Note that $$C_c(\mathbb R;\mathbb C)\subseteq\mathcal D(M_f).\tag1$$ Assume $f\in L^2(\lambda;\mathbb C)$. Let $g_0\in L^2(\lambda;\mathbb C)$ and $$Tg:=\langle f,g\rangle_{L^2(\lambda;\:\mathbb C)}g_0\;\;\;\text{for }g\in\mathcal D(M_f).$$ Note that $$\langle g,T^\ast h\rangle_{L^2(\lambda;\:\mathbb C)}=\langle Tg,h\rangle_{L^2(\lambda;\:\mathbb C)}=\left\langle g,\langle g_0,h\rangle_{L^2(\lambda;\:\mathbb C)}f\right\rangle_{L^2(\lambda;\:\mathbb C)}\tag2$$ for all $g\in\mathcal D(M_f)$ and $h\in\mathcal D(T^\ast)$.
Question 1: Usually I treat $L^2(\lambda;E)$ as the $\mathbb R$-vector space of $E$-valued ($E$ being any $\mathbb R$-Banach space) Bochner $\lambda$-integrable functions. However, since the context where I've encountered deals with $\mathbb C$-Hilbert spaces, I wonder whether we need to treat $L^2(\lambda;\mathbb C)$ as a $\mathbb C$-Hilbert space instead. Is this possible at all? And does it matter?
Note that I'm using the convention that a complex inner product is linear in the second argument and hence $(2)$ holds no matter whether $\langle\;\cdot\;,\;\cdot\;\rangle_{L^2(\lambda;\:\mathbb C)}$ is a complex or real inner product.
Question 2: I know that $C_c(\mathbb R)$ ($=\mathbb C_c(\mathbb R;\mathbb R)$) is dense in $L^2(\lambda)$ ($=L^2(\lambda;\mathbb R)$). But is $C_c(\mathbb R;\mathbb C)$ dense in $L^2(\lambda;\:\mathbb C)$ as well?
Question 3: How can we conclude that $$T^\ast h=\langle g_0,h\rangle_{L^2(\lambda;\:\mathbb C)}f\tag3$$ for all $h\in\mathcal D(T^\ast)$? I assume that the reasoning is that $\mathcal D(M_f)$ is dense, right? In that case, we are again by question 2, since the denseness of $C_c(\mathbb R;\mathbb C)$ would ensure the denseness of $\mathcal D(M_f)$.
Question 4: Why can we conclude that $$\langle g_0,h\rangle_{L^2(\lambda;\:\mathbb C)}=0\tag4$$ for all $h\in\mathcal D(T^\ast)$ and hence $\mathcal D(T^\ast)\perp g_0$? Moreover, why does it follow that $\mathcal D(T^\ast)$ is not dense and $\mathcal D(T^\ast)=(\mathbb C_0g_0)^\perp$?