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 V = W and we have B(v, w) = B(w, v) for all 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 (see 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 X, Y, 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 X, Y, and Z be locally convex Hausdorff spaces and let ${\displaystyle C:L\left(X;Y\right)\times L\left(Y;Z\right)\to L\left(X;Z\right)}$ be the composition map defined by ${\displaystyle C(u,v):=v\circ u}$. In general, the bilinear map 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 E is an equicontinuous subset of ${\displaystyle L\left(Y;Z\right)}$ then the restriction ${\displaystyle C{\big \vert }_{L\left(X;Y\right)\times E}:L\left(X;Y\right)\times E\to L\left(X;Z\right)}$ is continuous for all three topologies.[1]
• If Y is a barreled space then for every sequence ${\displaystyle \left(u_{i}\right)_{i=1}^{\infty }}$ converging to u in ${\displaystyle L\left(X;Y\right)}$ and every sequence ${\displaystyle \left(v_{i}\right)_{i=1}^{\infty }}$ converging to v in ${\displaystyle L\left(Y;Z\right)}$, the sequence ${\displaystyle \left(v_{i}\circ u_{i}\right)_{i=1}^{\infty }}$ converges to ${\displaystyle v\circ u}$ in ${\displaystyle L\left(Y;Z\right)}$. [1]