Bilinear Mappings and Multilinear Mappings

Bilinear Mappings and n-Linear 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:
(1)
\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:
(2)
\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:
(3)
\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 $x \in X$
(4)
\begin{align} \quad \sup_{y \in Y, \| y \| = 1} \| T_y(x) \| < \infty \end{align}
  • 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:
(5)
\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:
(6)
\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 $x_1 \in X_1$, $x_2 \in X_2$, …, $x_{i-1} \in X_{i-1}$, $x_{i+1} \in X_{i+1}$, …, $x_n \in X_n$ we have that the map $x \to T(x_1, x_2, ..., x_{i-1}, x, x_{i+1}, ..., x_n)$ 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.
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