1. The Abstract Problem
= a(v − σ f, v − σ f ) − a(σ f, σ f ).
The symmetry of a(·, ·) is essential in obtaining the last equality.
For a given f , since σ f is fixed, J is minimised if and only if a(v −
σ f, v − σ f ) is minimised. But this being the distance between v and
σ f , our knowledge of Hilbert space theory tells us that since K is a
closed convex subset, there exists a unique element u ∈ K such that this
minimum is obtained. This proves the existence and uniqueness of the
solution, which is merely the projection of σ f over K.
We know that this projection is characterised by the inequalities:
a(σ f − u, v − u) ≤ 0 for all
v ∈ K.
Geometrically, this means that the angle between the vectors (σ f −u)
and (v − u) is obtuse. See Fig. 1.1.
Thus, a(σ f, v − u) ≤ a(u, v − u) which by virtue of (1.7) is precisely
the relation (1.5). This completes the proof.
We can state the following
Corollary 1.1. (a) If K is a non-empty closed convex cone with vertex at origin 0, then the solution of (P) is characterised by:
a(u, v) ≥ f (v) for all v ∈ K
a(u, u) = f (u).