19.2. THE HILBERT SPACE L(U) 523

Note this is a good definition because {x : Lx = y} is closed thanks to the continuity ofL and it is obviously convex. Thus Theorem 19.1.8 applies. With this definition define aninner product on L(U) as follows. For y,z ∈ L(U) ,

(y,z)L(U) ≡(L−1y,L−1z

)U

The notation is abominable because L−1 (y) is the normal notation for My.

In terms of linear algebra, this L−1 is the Moore Penrose inverse. There you obtain theleast squares solution x to Lx = y which has smallest norm. Here there is an actual solutionand among those solutions you get the one which has least norm. Of course a real honestsolution is also a least squares solution so this is the Moore Penrose inverse restricted toL(U).

First I want to understand L−1 better. It is actually fairly easy to understand in terms ofgeometry. Here is a picture of L−1 (y) for y ∈ L(U).

Myw

L−1(y)

U

As indicated in the picture, here is a lemma which gives a description of the situation.

Lemma 19.2.2 In the context of the above definition, L−1 (y) is characterized by(L−1 (y) ,x

)U = 0 for all x ∈ ker(L)

L(L−1 (y)

)= y,

(L−1 (y) ∈My

)In addition to this, L−1 is linear and the above definition does define an inner product.

Proof: The point L−1 (y) is well defined as noted above. I claim it is characterized bythe following for y ∈ L(U)(

L−1 (y) ,x)

U = 0 for all x ∈ ker(L)

L(L−1 (y)

)= y,

(L−1 (y) ∈My

)Let w ∈My and suppose

(v,x)U = 0,L(v) = y