> [!quote] Introduction > > Given a set $X$, a ring $R$, and modules $E, F$ over $R$, it's often worthwhile to study the modules of functions from $X$ to these spaces. Given a linear map between these modules of functions, it's a natural question to ask whether it's possible to decompose it pointwise > $ > Tf(x) = T_x \cdot f(x) > $ > If we are simply given the space of all such functions, then it's easy to just choose $f$ to be an indicator function. However, usually the modules of interest do not form the space of *all* functions defined on $X$, but rather some strict subset (continuous, linear, differentiable, etc.). So the question we are given is actually kind of difficult. > [!definition] > > Let $R$ be a commutative [[Ring|ring]], $E, F$ be [[Module|modules]] over $R$, and $X$ be a [[Set|set]]. Define > - $E^X$ and $F^X$ as the module of functions from $X$ to $E$ and from $X$ to $F$. > - $R^X$ as the module of functions from $X$ to $R$. > > then $E^X$ is also a module over $R^X$. # Decomposition System > [!definition] > > Let $M_R \subset R^X$ and $M_E \subset E^X$ be sub-$R$-modules. The pair $(M_R, M_E)$ forms a **pointwise decomposition system** if the following conditions hold: > 1. $M_E$ is also a sub-$M_R$-module. > 2. **Constant Scalars:** For each $\alpha \in R$, there exists $\lambda \in M_R$ such that $\lambda(p) = \alpha$ for all $p \in X$. > 3. **Extension From Point**: For each $m \in E$ and $p \in X$, there exists $v \in M_E$ such that $v(p) = m$. > 4. **Zero Separation:** Let $p \in X$ and $v \in M_E$ such that $v(p) = 0$, then there exists a decomposition $v = \lambda_iv^i$ into a linear combination where $\lambda_i \in M_R$, $v^i \in M_E$, and $\lambda^i(p) = 0$ for all $i$. > [!theorem] > > Let $(M_R, M_E)$ be a pointwise decomposition system, and $T: M_E \to F^X$ be a $M_R$-[[Linear Transformation|linear map]], then for each $p \in X$, there exists $T_p \in L(E, F)$ such that > $ > Tv(p) = T_p v_p > $ > for all $v \in M_E$. > > *Proof*. Let $p \in X$, and define the map by > $ > T_p: E \to F \quad T_pm = (Tv)(p) > $ > where using condition $(3)$, $v$ is any element in $M_E$ such that $v(p) = m$. It's necessary to show that this map is well-defined. Let $w \in M_E$ such that $w(p) = m$ as well, then > $ > (Tv)(p) - (Tw)(p) = (Tv - Tw)(p) = T(v - w)(p) > $ > Thus it's only necessary to show that $(Tx)(p) = 0$ whenever $x(p) = 0$. To this end, let $x = \lambda_ix^i$ be a decomposition coming from condition $(4)$. Then by linearity, > $ > \begin{align*} > Tx = T(\lambda_ix^i) &= \lambda_iTx^i \\ > Tx(p) = T(\lambda_ix^i)(p) &= \lambda_i(p)\cdot Tx^i(p) = 0 > \end{align*} > $ > From here, the linearity of $T$ itself yields the desired results. > [!theorem] > > Let $(M_R, M_E)$ be a pointwise decomposition system, and $T: M_E^k \to F^X$ be a $k$-[[Multilinear Map|multilinear map]], then for each $p \in X$, there exists $T_p \in L^k(E, F)$ such that > $ > Tv(p) = T_pv_p > $ > *Proof*. Let $p \in X$, and define the map by > $ > T_p: E^k \to F \quad T_pm = [T(v_1, \cdots, v_k)](p) > $ > where $v_j(p) = m_j$ for each $1 \le j \le k$. It's necessary to show that this map is well-defined. Let $\seqf{w_j} \subset M_E$ such that $w_j(p) = m_j$ for each $1 \le j \le k$ as well. Then > $ > T(v_1, \cdots, v_k) - T(w_1, \cdots, w_k) > $ > has a multilinear decomposition > $ > \begin{align*} > &T(v_1, \cdots, v_k) - T(w_1, \cdots, w_k) \\ > &= \sum_{j = 1}^k T(v_1, \cdots, v_{j - 1}, v_j - w_j, w_{j + 1}, \cdots, w_k) > \end{align*} > $ > Evaluating at points, it's sufficient show that $T(x)(p) = 0$ whenever there exists $1 \le j \le k$ such that $x_j = 0$. To this end, apply the zero separation on such $x_j$ to yield the desired result.