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.

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 Bilnear 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 1: 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}

• 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:

(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 2: 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 Multilinear 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 1 and 2 above have analogues for multilinear maps.

Proposition 3: 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 4: 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.