Bilinear Mappings and Multilinear Mappings
Table of Contents

Bilinear Mappings and nLinear Mappings
Bilinear Mappings
Definition: Let $X$, $Y$, and $Z$ be linear spaces over $\mathbf{F}$. A function $T : X \times Y \to Z$ is said to be a Bilinear Map if for each fixed $y \in Y$, the restriction mapping $x \to T (x, y)$ is linear and if for each fixed $x \in X$, the restriction mapping $y \to T(x, y)$ is linear. 
We first translate the definition above to some basic properties of bilinear maps.
Proposition 1: Let $X$, $Y$, and $Z$ be linear spaces over $\mathbf{F}$ and let $T : X \times Y \to Z$ be a bilinear map. Then: a) $T(x_1 + x_2, y) = T(x_1, y) + T(x_2, y)$ for all $x_1, x_2 \in X$ and for all $y \in Y$. b) $T(x, y_1 + y_2) = T(x, y_1) + T(x, y_2)$ for all $x \in X$ and for all $y_1, y_2 \in Y$. c) $T(\alpha x, y) = \alpha T(x, y) = T(x, \alpha y)$ for all $x \in X$, for all $y \in Y$, and for all $\alpha \in \mathbf{F}$. d) If $x = 0_X$ or $y = 0_X$ then $T(x, y) = 0_Z$. 
 Proof of a) Let $x_1, x_2 \in X$, $y \in Y$. Then since $x \to T(x, y)$ is linear we have that $T(x_1 + x_2, y) = T(x_1, y) + T(x_2, y)$. $\blacksquare$
 Proof of b) Let $x \in X$, $y_1, y_2 \in Y$. Then since $y \to T(x, y)$ is linear we have that $T(x, y_1 + y_2) = T(x, y_1) + T(x, y_2)$. $\blacksquare$
 Proof of c) Let $x \in x$, $y \in Y$, $\alpha \in \mathbf{F}$. Since $y \to T(x, y)$ is linear we have that $T(\alpha x, y) = \alpha T(x, y)$, and since $x \to T(x, y)$ is linear we have that $T(x, \alpha y) = \alpha T(x, y)$. Thereefore $T(\alpha x, y) = \alpha T(x, y) = T(x, \alpha y)$. $\blacksquare$
 Proof of d) If $x = 0_X$ then by (c) we have that $T(0_X, y) = T(00_X, y) = 0T(0_X, y) = 0_Z$. Similarly, if $y = 0_Y$ then we have that $T(x, 0_Y) = T(x, 00_Y) = 0T(x, 0_Y) = 0_Z$. $\blacksquare$
Definition: Let $X$, $Y$, and $Z$ be normed linear spaces over $\mathbf{F}$. A bilinear map $T : X \times Y \to Z$ is said to be bounded if there exists an $M > 0$ such that $\ T (x, y) \_Z \leq M \ x \_X \ y \_Y$ for all $x, y \in X$. If $T$ is bounded then the Operator Norm of $T$ is defined to be $\displaystyle{\ T \ = \sup_{\ x \_X \leq 1, \ y \_Y \leq 1} \{ \ T(x, y) \ \}}$. The Set of All Bounded Linear Maps from $X \times Y$ to $Z$ is denoted $\mathrm{BL}(X, Y; Z)$. 
When it is not ambiguous we will denote the norm of $X$ and the norm of $Y$ simply by "$\ \cdot \$".
If $X$, $Y$, and $Z$ are Banach spaces over $\mathbf{F}$ then as we will see in the following proposition, a bilinear map $T: X \times Y \to Z$ will be bounded whenever $T$ is separately continuous.
Proposition 2: Let $X$, $Y$ and $Z$ be Banach spaces over $\mathbf{F}$ and let $T : X \times Y \to Z$ be a bilinear map. If $T$ is separately continuous then $T$ is bounded. 
In this context, a bilinear map $T : X \times Y \to Z$ is separately continuous if for each fixed $y_0 \in Y$ we have that $x \to T(x, y_0)$ is continuous and for each fixed $x_0 \in X$ we have that $y \to T(x_0, y)$ is continuous.
 Proof: For each fixed $x_0 \in X$ let $T_{x_0} : Y \to Z$ be defined for all $y \in Y$ by:
\begin{align} \quad T_{x_0}(y) = T(x_0, y) \end{align}
 And for each fixed $y_0 \in Y$ let $T_{y_0} : X \to Z$ be defined for all $x \in X$ by:
\begin{align} \quad T_{y_0}(x) = T(x, y_0) \end{align}
 Since $T$ is separately continuous we have that $T_{x_0}$ and $T_{y_0}$ are continuous (and hence bounded) for each $x_0 \in X$ and for each $y_0 \in Y$.
 So for each $x \in X$, since $T_x$ is bounded there exists an $M_x > 0$ such that $\ T_x(y) \ \leq M_x \ y \$ for every $y \in Y$. Therefore:
\begin{align} \quad \sup_{y \in Y, \ y \ = 1} \ T_y(x) \ = \sup_{y \in Y, \ y \ = 1} \ T(x, y) \ = \sup_{y \in Y, \ y \ = 1} \ T_x(y) \ \leq \sup_{y \in Y, \ y \=1} M_x \ y \ = M_x < \infty \end{align}
 Now observe that $\mathcal F = \{ T_y : y \in Y, \ y \ =1 \}$ is a collection of bounded linear operators from the Banach space $X$ to the Banach space $Z$. Furthermore, from above we have that for each $x \in X$
\begin{align} \quad \sup_{y \in Y, \ y \ = 1} \ T_y(x) \ < \infty \end{align}
 Therefore by The Uniform Boundedness Principle we have that $\displaystyle{\sup_{y \in Y, \ y \=1} \ T_y \ = M < \infty}$.
 Now suppose that $y = 0$. Then observe that $\ T(x, y) \ = \ T(x, 0) \ = \ T_x(0) \ = 0$ for all $x \in X$. So let $x \in X$, $y \in Y$ be such that $y \neq 0$. Then $\left \ \frac{y}{\ y \} \right \ = 1$ and so:
\begin{align} \quad \frac{1}{\ y \} \ T(x, y) \ = \left \ T \left ( x, \frac{y}{\ y \} \right ) \right \= \ T_{y/\ y \} (x) \ = \leq M \ x \ \end{align}
 Rearranging the above equation and we see that for all $x \in X$ and for all $y \in Y$ we have that:
\begin{align} \quad \ T(x, y) \ \leq M \ x \ \ y \ \end{align}
 So $T$ is bounded. $\blacksquare$
Proposition 3: Let $X$, $Y$, and $Z$ be normed linear spaces over $\mathbf{F}$. Then $\mathrm{BL}(X, Y; Z)$ with the operations of function addition, scalar multiplication, and with norm defined for all $T \in \mathrm{BL}(X, Y; Z)$ by $\ T \ = \sup_{\ x \ \leq 1, \ y \ \leq 1} \{ \ T(x, y) \ \}$ is a normed linear space. Furthermore, if $Z$ is a Banach space then $\mathrm{BL}(X, Y; Z)$ is a Banach space. 
Multilinear Mappings
We can also define multilinear mappings as follows.
Definition: Let $X_1$, $X_2$, …, $X_n$, $Z$ be linear spaces. A mapping $T : X_1 \times X_2 \times ... \times X_n \to Z$ is said to be an $n$Multilinear Map if for each $1 \leq i \leq n$ and for all fixed $x_1 \in X_1$, $x_2 \in X_2$, …, $x_{i1} \in X_{i1}$, $x_{i+1} \in X_{i+1}$, …, $x_n \in X_n$ we have that the map $x \to T(x_1, x_2, ..., x_{i1}, x, x_{i+1}, ..., x_n)$ is linear. 
Definition: Let $X_1$, $X_2$, …, $X_n$, $Z$ be normed linear spaces and let $T : X_1 \times X_2 \times ... \times X_n \to Z$ be an $n$multilinear map. Then $T$ is said to be Bounded if there exists an $M > 0$ such that $\ T(x_1, x_2, ..., x_n) \_Z \leq M \ x_1 \_{X_1} \ x_2 \_{X_2} ... \ x_n \_{X_n}$ for all $x_i \in X_i$, $1 \leq i \leq n$. If $T$ is bounded then the Operator Norm of $T$ is defined to be $\displaystyle{\ T \ = \sup_{\ x_1 \_{X_1}, \ x_2 \_{X_2}, ..., \ x_n \_{X_n} \leq 1} \{ \ T(x_1, x_2, ..., x_n) \ \}}$. The Set of All Bounded Linear Operators from $X_1 \times X_2 \times ... \times X_n$ to $Z$ is denoted $\mathrm{BL}(X_1, X_2, ..., X_n; Z)$. 
Propositions 2 and 3 above have analogues for multilinear maps.
Proposition 4: Let $X_1$, $X_2$, …, $X_n$ and $Z$ be Banach spaces over $\mathbf{F}$ and let $T : X_1 \times X_2 \times ... \times X_n \to Z$ be a bilinear map. If $T$ is separately continuous then $T$ is bounded. 
Proposition 5: Let $X_1$, $X_2$, …, $X_n$ and $Z$ be normed linear spaces over $\mathbf{F}$. Then $\mathrm{BL}(X_1, X_2, ..., X_n; Z)$ with the operations of function addition, scalar multiplication, and with norm defined for all $T \in \mathrm{BL}(X, Y; Z)$ by $\ T \ = \sup_{\ x_1 \_{X_1}, \ x_2 \_{X_2}, ..., \ x_n \_{X_n} \leq 1} \{ \ T(x_1, x_2, ..., x_n) \ \}$ is a normed linear space. Furthermore, if $Z$ is a Banach space then $\mathrm{BL}(X_1, X_2, ..., X_n; Z)$ is a Banach space. 