(1) Could anyone tell me how to find a Dual space of the following space of continuous functions of the following form? And dual maps?
$V=\{f: K\to \mathbb R^n, f(x)=Ax+b\}$, where $K$ is a compact subset of $\mathbb R^n$ and $A$ is a matrix of order $n\times n$, $b\in \mathbb R^n$ a $n\times 1$ vector.
What is $V^*$? How to apply the Riesz-Representation theorem here? What would a typical dual map look like? Thanks a lot.
What if $K$ is a closed subset only?
(2) $T: C_b(\mathbb R^n,\mathbb R)\to C_b(\mathbb R^n,\mathbb R), T(f)= \sum_{i=1}^{n} p_i (f\circ f_i)(x)$ where $f_i\in V$, $T$ is a linear map. Could any one tell me what is $T^*$?
The standard Riesz representation theorem is for continuous real-valued functions, whereas here we have $\mathbb{R}^n$-valued functions. However, this is not a huge hurdle as we can essentially apply everything component-wise.
Let $\mu$ be a vector-valued Borel measure on $K$. Then, for every Borel set $B \subset K$, $$\mu(B) = (\mu_1(B), \mu_2(B), \ldots, \mu_n(B)) \in \mathbb{R}^n,$$ where each $\mu_i$ is a signed (that is, real-valued) Borel measure on $K$.
We first apply the Riesz representation on $C(K; \mathbb{R}^n)$. Therefore, for every $f \in C(K; \mathbb{R}^n)$, a linear functional $\lambda$ looks like $$ \lambda(f) = \int_K \langle f(x), d\mu(x)\rangle$$ where $\langle f(x), d\mu(x) \rangle = f_1(x) \, d\mu_1(x) + \ldots + f_n(x) \, d\mu_n(x)$ and $\mu$ is a vector-valued Borel measure.
So our next step is to construct $V'$, where $V$ is the set of all linear functions on $K$. Since $f(x) = Ax + b$, where $A$ is $n \times n$ and $b$ is $n \times 1$, then it follows that $V$ is an $(n^2 + n)$-dimensional space; thus it must also be true that $V'$ is an $(n^2 + n)$-dimensional space.
Recall that when we want to construct the dual space of a finite-dimensional vector space, the easiest way to do this is to construct a dual basis. Recall that if $\{v_1, \ldots, v_n\}$ is the basis for $V$, then the dual basis $\{f_1, \ldots, f_n\}$ is such that $$f_j(v_i) = \delta_{ij}.$$
Since every $f \in V$ is of the form $f(x) = Ax + b$, an easy basis for $V$ consists of $n^2$ vectors that correspond to the "$Ax$" term, and then $n$ vectors that correspond to the "$b$" term. In particular, we can choose a basis $\{E_{11}x, \ldots, E_{nn}x, E_1, \ldots, E_n\}$, where $E_{ij}$ is the $n \times n$ matrix which has all its entries equal to zero except the $(i,j)$ component, and $E_i$ is a constant function whose constant value is a vector in $\mathbb{R}^n$ whose components are all zero except the $i$th component.
To construct the dual vectors, it is instructive to look at the one-dimensional case; that is $n = 1$. Then $V$ is two-dimensional, spanned by the basis $\{x, 1\}$. Then a dual basis consists of two signed measures $\mu, \nu$, where \begin{equation} \int_K x \, d\mu = 1, \int_K 1 \, d\mu = 0 \\ \int_K x \, d\nu = 0, \int_K 1 \, d\nu = 1. \end{equation} Now, let $k_1, k_2 \in K$ be distinct points in $K$, and let $\delta_{k_1}, \delta_{k_2}$ be the Dirac masses at these two points. One can check that \begin{equation} \mu = \frac{1}{k_1 - k_2} \delta_{k_1} - \frac{1}{k_1 - k_2} \delta_{k_2} \\ \nu = \frac{-k_2}{k_1 - k_2} \delta_{k_1} + \frac{k_1}{k_1 - k_2} \delta_{k_2} \end{equation}
satisfies these two conditions. It then remains to extend this result, in a componentwise manner, to construct the $n^2 + n$ dual vectors in the general case.
It is important to note also that while the Riesz theorem says that every linear functional on $C(K)$ is given by a unique signed measure, in this exercise we aren't looking at the dual of $C(K)$, but rather the dual of $V$, which is a finite-dimensional space. Therefore, while each linear functional can be represented by a measure, that measure is not unique. There will be many different measures whose corresponding linear functional behaves identically when restricted only to $V$.
Furthermore, in light of this, it is apparent that we don't need $K$ to be a compact set anymore.