Linear transformations inside the convex algebra #
A linear transformation is a single-valued convex process, and ConvexProcess.ofLinearMap is that
embedding. Every operation of the convex algebra restricts along it to the corresponding operation
on linear maps: the image of a set is the linear image, composition is composition, the image of a
function under the associated bifunction is mapLin, and the adjoint of the process is a transpose
of the linear map.
Implementation notes #
The adjoints of a convex process, bifunction and function are defined outright from a pair of
pairings; none mentions a linear map. A transpose enters only here, as the hypothesis
IsAdjointPair Bu Bx T T' — between arbitrarily paired spaces a linear map need not have a
transpose at all, and when it does it is unique only if Bu is right-separating.
adjointProcess and coadjointProcess must be kept apart on a general process, but on a linear one
the graph is a subspace, so the defining inequality at -u reverses the one at u and both
collapse to ⟨T u, y⟩ = ⟨u, v⟩. It is right-separation of Bu, not of Bx, that pins the answer:
without it the adjoint process is single-valued only up to the annihilator of U in V.
References #
- R. T. Rockafellar, Convex Analysis, Princeton University Press, 1970, §39.
The set-level dictionary #
The image of a function under a linear process #
Ff at the indicator bifunction of a linear T is the image mapLin T f. The hypothesis
that f is nowhere ⊥ is not a convenience: off the fibre of T the summand is ⊤, and
⊥ + ⊤ = ⊥ would drag the infimum to ⊥ at every point of X.
The adjoint of a linear process #
The adjoint of a linear process is a transpose of the linear map. Bu.SeparatingRight is
what makes the answer unique, as it is for the transpose itself.
The infimum-oriented adjoint of a linear process is the same transpose, the defining inequality holding in both directions on a subspace.