R¦ L L0 , ¦L R¦ L0

T

T

F¦ L (©) FT

0

(4) T ’1 : FT ’ F L ( ) coincides with L. To see this, choose an arbitrary f = (» —¦

0

L) y (·, y) ∈ FT , with » ∈ L0 . Then we have T ’1 f = » y L (·, y), leading to

0

T ’1 f (x) = » y (x, y)

= » y (δx —¦ L)u (δ y —¦ L)v (u, v)

= (δx —¦ L)u (» —¦ L)v (u, v)

= (δx —¦ L)u f (u)

= L f (x).

(5) Since both T and T — map dense subspaces of Hilbert spaces, into Hilbert spaces, they

possess unique isometric extensions

( )— ’ LT := T — (N ( )— ) ⊆ N ( )— ,

T— : N L L

T :N ( ) ’ FT := T (N ( )) ⊆ N ( ).

L L

’1

Since both T —¦ R L —¦ T — and R |LT , are isometric extensions of R |L0 , they have to

T

coincide. Thus, the following diagram has to be commutative also:

˜

T*

N¦ L (©)* LT

R¦ LT

R¦ L

˜

T

N¦ L (©) FT

The next step is to show that both the extensions T and T — are surjective, i.e. that

LT = N ( )— and FT = N ( ), respectively.

300 Generalized interpolation

(6) Since N L ( ) is complete and since T : N L ( ) ’ FT ⊆ N ( ) is an isometric

isomorphism, FT has also to be complete. But this means that FT is a reproducing-

kernel Hilbert function space with reproducing kernel . Thus, by Theorem 10.11 we

have FT = N ( ). Moreover, if we have an arbitrary » ∈ N ( )— then the Riesz mapping

gives R (») = » y (·, y) ∈ N ( ) = FT and therefore (R |LT )’1 R (») ∈ LT , showing

that R |LT = R or, in other words, LT = N ( )— .

The equality FT = N ( ) has an important side effect. The space FT is the image of

N L ( ) under the mapping T , which is the extension of the mapping T from the dense

subspace F L ( ) of N L ( ). Hence, T (F L ( )) = FT is dense in N ( ).

0

(7) The density mentioned in the last point allows us to show that T ’1 = L, as follows. If

f ∈ N ( ) is given then we can choose a sequence f n = (»n —¦ L) y (·, y) ∈ FT , »n ∈ L0 ,

0

with f ’ f n N ( ) ’ 0 for n ’ ∞. Since δx —¦ L ∈ N ( )— this means that on the one

hand

|L f (x) ’ L f n (x)| ¤ δx —¦ L f ’ fn ’ 0, n ’ ∞.

N ( )— N()

On the other hand, we know by step (4) that T ’1 f n (x) = L f n (x), giving

|T ’1 f (x) ’ L f n (x)| ¤ δx —¦ T ’1 f ’ fn ’ 0, n ’ ∞.

N ( )— N ( )—

This establishes the identity L = T ’1 . Thus for every f ∈ N ( ) we have L f = T ’1 f ∈

N L ( ) and L f N L ( ) = f N ( ) .

(8) Similarly, we can establish the identity T — (») = » —¦ L for all » ∈ N L ( )— . Namely,

for any » ∈ N L ( )— there exists owing to step (1) a sequence {»n } ⊆ L0, L approximating

» for n ’ ∞. Thus if f ∈ N ( ) is given then we know by step (7) that L f ∈ N L ( ),

showing that

|» —¦ L( f ) ’ »n —¦ L( f )| ¤ » ’ »n ’ 0, n ’ ∞.

Lf

( )—

N N ()

L L

By de¬nition we have T — (») = » —¦ L so that

|T — (»)( f ) ’ »n —¦ L( f )| ¤ T — (» ’ »n ) f

N ( )— N()

= » ’ »n ’ 0, n ’ ∞.

f

N ( )— N()

( )— we now know that » —¦ L =

This allows us to conclude the proof. For » ∈ N L

— (») ∈ N ( )— and » —¦ L

N ( )— = » N L ( )— .

T

After discussing the relations between the native spaces of and we now investigate

L

the connection between their power functions.

( )— and » ∈ N ( )— are given.

= {»1 , . . . , » N } ⊆ N

Theorem 16.10 Suppose that L L

Then

(») = P —¦ L).

P , —¦L (»

L,

16.3 Solving PDEs by collocation 301

Proof By de¬nition and by Theorem 16.9 we have that

—¦ L) = »—¦L ’μ

P , —¦L (» inf N ( )—

μ∈span{ —¦L}

= »—¦L ’μ—¦L

inf N ( )—

μ∈span{ }

= »’μ

inf ( )—

N

μ∈span{ } L

=P (»).

L,

Our next result also deals with the power function. This time we want to investigate what

happens if we drop some of the functionals. The result should no longer astonish us.

= {»1 , . . . , » N } ⊆ N ( )— and ⊆

Theorem 16.11 Suppose that are given. Then

(») ¤ P

P (»)

, ,