Bilinear map

Let ${\displaystyle V,W}$ and ${\displaystyle X}$ be three vector spaces over the same base field ${\displaystyle F}$. A bilinear map is a function

$${\displaystyle B:V\times W\to X}$$

such that for all ${\displaystyle w\in W}$, the map ${\displaystyle B_{w}}$

$${\displaystyle v\mapsto B(v,w)}$$

is a linear map from ${\displaystyle V}$ to ${\displaystyle X,}$ and for all ${\displaystyle v\in V}$, the map ${\displaystyle B_{v}}$

$${\displaystyle w\mapsto B(v,w)}$$

is a linear map from ${\displaystyle W}$ to ${\displaystyle X.}$ In other words, when we hold the first entry of the bilinear map fixed while letting the second entry vary, the result is a linear operator, and similarly for when we hold the second entry fixed.

Such a map ${\displaystyle B}$ satisfies the following properties.

• For any ${\displaystyle \lambda \in F}$, ${\displaystyle B(\lambda v,w)=B(v,\lambda w)=\lambda B(v,w).}$
• The map ${\displaystyle B}$ is additive in both components: if ${\displaystyle v_{1},v_{2}\in V}$ and ${\displaystyle w_{1},w_{2}\in W,}$ then ${\displaystyle B(v_{1}+v_{2},w)=B(v_{1},w)+B(v_{2},w)}$ and ${\displaystyle B(v,w_{1}+w_{2})=B(v,w_{1})+B(v,w_{2}).}$

If ${\displaystyle V=W}$ and we have B(v, w) = B(w, v) for all ${\displaystyle v,w\in V,}$ then we say that B is symmetric. If X is the base field F, then the map is called a bilinear form, which are well-studied (for example: scalar product, inner product, and quadratic form).

The definition works without any changes if instead of vector spaces over a field F, we use modules over a commutative ring R. It generalizes to n-ary functions, where the proper term is multilinear.

For non-commutative rings R and S, a left R-module M and a right S-module N, a bilinear map is a map B : M × N → T with T an (R, S)-bimodule, and for which any n in N, m ↦ B(m, n) is an R-module homomorphism, and for any m in M, n ↦ B(m, n) is an S-module homomorphism. This satisfies

B(r ⋅ m, n) = r ⋅ B(m, n)

B(m, n ⋅ s) = B(m, n) ⋅ s

for all m in M, n in N, r in R and s in S, as well as B being additive in each argument.

Many bilinear maps that occur in practice are separately continuous but not all are continuous. We list here sufficient conditions for a separately continuous bilinear to be continuous.

• If X is a Baire space and Y is metrizable then every separately continuous bilinear map ${\displaystyle b:X\times Y\to Z}$ is continuous.[1]
• If ${\displaystyle X,Y,{\text{ and }}Z}$ are the strong duals of Fréchet spaces then every separately continuous bilinear map ${\displaystyle b:X\times Y\to Z}$ is continuous.[1]
• If a bilinear map is continuous at (0, 0) then it is continuous everywhere.[2]

Let ${\displaystyle X,Y,{\text{ and }}Z}$ be locally convex Hausdorff spaces and let ${\displaystyle C:L(X;Y)\times L(Y;Z)\to L(X;Z)}$ be the composition map defined by ${\displaystyle C(u,v):=v\circ u.}$ In general, the bilinear map ${\displaystyle C}$ is not continuous (no matter what topologies the spaces of linear maps are given). We do, however, have the following results:

Give all three spaces of linear maps one of the following topologies:

1. give all three the topology of bounded convergence;
2. give all three the topology of compact convergence;
3. give all three the topology of pointwise convergence.

• If ${\displaystyle E}$ is an equicontinuous subset of ${\displaystyle L(Y;Z)}$ then the restriction ${\displaystyle C{\big \vert }_{L(X;Y)\times E}:L(X;Y)\times E\to L(X;Z)}$ is continuous for all three topologies.[1]
• If ${\displaystyle Y}$ is a barreled space then for every sequence ${\displaystyle \left(u_{i}\right)_{i=1}^{\infty }}$ converging to ${\displaystyle u}$ in ${\displaystyle L(X;Y)}$ and every sequence ${\displaystyle \left(v_{i}\right)_{i=1}^{\infty }}$ converging to ${\displaystyle v}$ in ${\displaystyle L(Y;Z),}$ the sequence ${\displaystyle \left(v_{i}\circ u_{i}\right)_{i=1}^{\infty }}$ converges to ${\displaystyle v\circ u}$ in ${\displaystyle L(Y;Z).}$ [1]