> [!definition] > > Let $\seqf{X_k}$ be a [[Sequence|sequence]] of [[Vector Space|vector spaces]], $X = \bigotimes_{k = 1}^{n}X_k$ be their [[Direct Sum|direct sum]], and $Y$ be another vector space. A map $\lambda: X \to Y$ is **multilinear** if it is [[Linear Transformation|linear]] in each of the coordinates of $X$. > [!definition] > > Let $X, Y$ be a vector space and > $ > \lambda: \bigotimes_{k = 1}^{n}X \to Y > $ > be a multilinear map. $\lambda$ is **symmetric** if for any $\sigma \in S_n$, $\lambda \circ \sigma = \lambda$. In other words, the [[Symmetric Group|permutation]] of its arguments do not affect its value. > [!theorem] > > Let $R$ be a commutative [[Ring|ring]], $\seqf{E_j}$ and $F$ be $R$-modules, and $\lambda: \prod_jE_j \to F$ be a multilinear map. Let $x, y \in \prod_j E_j$, then > $ > \lambda(x) - \lambda(y) = \sum_{j = 1}^n \lambda(x_1, \cdots, x_{j - 1}, x_j - y_j, a_{j + 1}, \cdots, a_n) > $ > *Proof*, by induction on $n$. The base case is direct. Suppose that the proposition holds for $n$, then > $ > \begin{align*} > \lambda(x) - \lambda(y) &= \lambda(x_{\ol{n}}, x_{n + 1}) - \lambda(y_{\ol{n}}, y_{n + 1}) \\ > &= \lambda(x_{\ol n} - y_{\ol{n}}, x_{n + 1}) + \lambda(y_{\ol{n}}, x_{n + 1}) \\ > &- \lambda(y_{\ol{n}}, y_{n + 1} - x_{n + 1}) - \lambda(y_{\ol{n}}, x_{n + 1}) \\ > &= \lambda(x_{\ol n} - y_{\ol{n}}, x_{n + 1}) + \lambda(y_{\ol{n}}, x_{n + 1} - y_{n + 1}) > \end{align*} > $ > For fixed $x_{n + 1}$, the inductive hypothesis can be applied to yield the desired result. # Bounded Multilinear Maps > [!definition] > > Let $\seqf{E_j}$ and $F$ be [[Banach Space|Banach spaces]]. Let $\lambda: \prod_j E_j \to F$ be a [[Multilinear Map|multilinear map]], then $\lambda$ is **bounded** if there exists $C \ge 0$ such that > $ > \norm{\lambda x} \le C \prod_{j}\norm{x_j} > $ > where the infimum over all such $C$ is the **norm** of $\lambda$, which makes the space $L^n(\seqf{E_j}, F)$ a Banach space. > [!theorem] > > Let $E$ and $F$ be Banach spaces, and let $L^r(E, F)$ be the space of all bounded $r$-linear maps from $\prod_{j = 1}^r E$ to $F$. Then if we fix $F$, > 1. $L^r$ is a contravariant [[Functor|functor]]. > 2. The mapping $L(E_1, E_2) \to L(L^r(E_2, F), L^r(E_1, F))$ is [[Space of Smooth Functions|smooth]]. > > *Proof*. Let $T \in L(E_1, E_2)$, and define the mapping > $ > L^r(T): L^r(E_2, F) \to L^r(E_1, F) \quad \lambda \mapsto \lambda \circ T > $ > where $T$ is composed at each argument. Note that $L^r$ as a functor admits an extension to $L(E_1, E_2)^r$ via composition at each argument. Here, $L^r$ is a bounded multilinear map, and hence is smooth. Therefore its restriction is also smooth. > [!theorem] > > Let $\cx_1, \cx_2, \cy$ be [[Normed Vector Space|normed spaces]], $\lambda: \cx_1 \times \cx_2 \to \cy$ be a bilinear map, and $\psi: \cx_1 \times \cx_2 \to \cy$ defined on a [[Neighbourhood|neighbourhood]] $U \in \cn(0)$ such that > $ > \lim_{(x_1, x_2) \to 0}\psi(x_1, x_2) = 0 > $ > and > $ > \norm{\lambda(x_1, x_2)} \le \norm{\psi(x_1, x_2)} \cdot \norm{x_1} \cdot \norm{x_2} > $ > then $\lambda = 0$. > > *Proof*. Let $(x_1, x_2) \in \cx_1 \times \cx_2$, and $t > 0$ such that $t(x_1, x_2) \in U$, then > $ > \begin{align*} > \norm{\lambda(tx_1, tx_2)} &\le \norm{\psi(t x_1, t x_2)} \cdot \norm{t x_1} \cdot \norm{t x_2} \\ > t^2\norm{\lambda(x_1, x_2)} &\le t^2 \cdot \norm{\psi(t x_1, t x_2)} \cdot \norm{x_1} \cdot \norm{x_2} \\ > \norm{\lambda(x_1, x_2)} &\le \norm{\psi(t x_1, t x_2)} \cdot \norm{x_1} \cdot \norm{x_2} > \end{align*} > $ > where sending $t \to 0$ yields that $\norm{\lambda(x_1, x_2)} = 0$. > [!theorem] > > Let $\seqf{X_k}$ be a sequence of vector spaces, $X = \bigotimes_{k = 1}^{n}X_k$ be their product, $Y, Z$ be vector spaces. If $\omega: X \to Y$ is a continuous multilinear map, and $\lambda \in L(Y, Z)$ be a [[Bounded Linear Map|bounded linear map]], then > 1. $\lambda \circ \omega$ is also a continuous multilinear map. > 2. The composition map $\lambda_*$ with $\omega \mapsto \lambda \circ \omega$ is a bounded linear map. > > ### Composition > > Let $f \in \prod_{k = 1}^{n}X_k$, then > $ > \begin{align*} > \norm{\lambda \circ \omega \circ f} &\le \norm{\lambda} \cdot \norm{\omega \circ f} \\ > &\le \norm{\lambda} \cdot \norm{\omega} \cdot \prod_{k = 1}^{n}\norm{f(k)} > \end{align*} > $ > Therefore the composition is bounded. > > ### Composition Map > > Let $\omega_1, \omega_2$ be multilinear maps and $\alpha$ be a scalar, then > $ > \begin{align*} > \lambda(\alpha \omega_1 + \omega_2)(x) &= \lambda (\alpha \omega_1(x) + \omega_2(x)) \\ > &= \alpha \lambda \omega_1(x) + \lambda \omega_2(x) > \end{align*} > $ > the composition map is linear. > > Let $\omega$ be a multilinear map, then > $ > \norm{(\lambda \circ \omega)(x)} \le \underbrace{\norm{\lambda} \cdot \norm{\omega}}_{\ge \norm{\lambda \circ \omega}} \cdot \prod_{k = 1}^{n}\norm{x_k} > $ > we have $\norm{\lambda \circ \omega} \le \norm{\lambda} \cdot \norm{\omega}$. Therefore the composition map is continuous.