Saito Frameformula

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An approximation formula in Hilbert space

Zhihua Zhang and Naoki Saito

Dept. of Math., Univ. of California, Davis, California, 95616, USA.

E-mail: [email protected] and [email protected]

Abstract Let {k}

1 be a frame for Hilbert space H. The purpose of this paper is to present an approximation

formula of any f H by a linear combination of finitely many frame elements in the frame {k}

1 and show that the

obtained approximation error depends on the bounds of frame and the convergence rate of frame coefficients of f as well

as the relation among frame elements.

Key words: approximation formula; linear combination; frame element; Hilbert space

MSC 42C15

1. Introduction

Let H be a Hilbert space and {ek}1 be an orthonormal basis for H. It is well-known that any f H canbe approximated by the linear combination of {ek}n1 and the approximation error depends on the convergence

rate of the Fourier coefficients[2].

As a generalization of the orthonormal bases, Duffin and Schaeffer[3] introduced the notion of frames.

Suppose that {k}1 is a frame for H. For any f H, we will construct a linear combination of finitely many

frame elements in the frame {k}1 to approximate to f and show that the approximation error depends on

the bounds of frame and the convergence rate of frame coefficients of f as well as the relation among frame

elements.

We recall some concepts and propositions.

The work was supported by NSF grant DMS-0410406 and Prof. Xiaoping Shens NSF grant

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Let {k}1 be a sequence in Hilbert space H. If there exist two positive constants A, B such that

A f 2k=1

|(f, k)|2 B f 2 f H, (1.1)

then the sequence {k}1 is said to be a frame for H, where A and B are said to be frame bounds. Specially,

if A = B = 1 and k = 1 (k Z+), then {k}1 is an orthonormal basis for H.

Let {k}1 be a frame for H. The frame operator S is defined as

S : H H, Sf =k=1

(f, k)k f H, (1.2)

where (f, k) (k Z+) are said to be frame coefficients.

Proposition 1.1[1,p58-62]. Let {k}1 be a frame with bounds A and B for H. Then

(i) The frame operator S is a self-conjugate operator and A f Sf B f f H.

(ii) The inverse operator S1 exists and 1B

f S1f 1A

f f H.

(iii) Denote k = S1k, the {k}1 is also a frame for H and1

B f

2

k=1 |(f, k)|2 1

A f

2

f

H. (1.3)

(iv) For each k,

k = 2A + B

k +2

A + B

n=1

I 2S

A + B

nk, (1.4)

where I is the identity operator.

(v) Let R = I 2SA+B

. Then R BAB+A

.

The frame {k}1 is said to be a dual frame of {k}1 .

Denote the partial sum of the series in (1.4) by Nk , i.e.0k = 2A + B k, Nk = 2A + B k + 2A + B

Nn=1

I 2S

A + B

nk. (1.5)

Proposition 1.2[1,p58-62]. Under the conditions of Proposition 1.1, then

(i) for any f H, the reconstruction formula f =k=1

(f, k)k holds2


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(ii) for any f H, the seriesk=1

(f, k)

Nk is convergent and

f k=1

(f, k)Nk qN+1 f (N Z+), (1.6)where q = BA

B+A.

2. A new frame approximation operator (Nn ())m

In order to approximate to any f H by a linear combination of finitely many frame elements, we present

a new frame approximation operator in this section.

Let

{k

}1 be a frame for H with bounds A and B. The frame operator S, k, and

Nk are stated in

(1.2), (1.4), and (1.5), respectively.

Definition 2.1 We truncate the seriesk=1

(f, k)Nk by its partial sum, for n, N Z+, defineNn (f) :=

nk=1

(f, k)Nk f H. (2.1)Denote

Sm : H H, Smf :=m

j=1(f, j)j f H. (2.2)

and

S1m := Sm, Slm := S

l1m (Sm) (l Z+),

S1 := S, Sl := Sl1(S) (l Z+). (2.3)

Definition 2.2. For any k , N, m Z+, define

(

0k)m =

2

A + Bk and (

Nk )m =

2

A + Bk +

2

A + B

Nn=1

I 2Sm

A + B

nk.

From this definition, we have

(Nk )m = (N1k )m + 2A + B

I 2SmA + B

Nk (N Z+). (2.4)

Definition 2.3. Define a frame approximation operator as follows.

(Nn ())m : H H, (Nn (f))m =n

k=1

(f, k)(Nk )m f H.3


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Lemma 2.4. (i) The sequence of operators {Sm}1 is uniformly bounded.

(ii) For any f

H,

Slmf Slf (m ) l Z+. (2.5)

Proof. Combining (1.2) with (2.2), for any f H, we have

Smf Sf (m ). (2.6)

Using the resonance theorem, we know that there exists M > 0 such that Sm M (m Z+). So we get (i).

We have known that (2.5) holds for l = 1. Now we assume that (2.5) holds for l

1.

Noticing that Slmf Slf = Sm(Sl1m f) Sm(Sl1f) + Sm(Sl1f) S(Sl1f), we have

Slmf Slf Sm(Sl1m f Sl1f) + Sm(Sl1f) S(Sl1f) =: p(l)m (f) + q(l)m (f). (2.7)

Since Sm M, by the postulate of induction, we have

p(l)m (f) M Sl1m f Sl1f 0 (m ).

Let g = Sl1f. Then by (2.6),

q(l)m (f) = Smg Sg 0 (m ).

Hence, by (2.7), we have Slmf Slf (m ), i.e. (2.5) holds for any l Z+. So we get (ii).

Lemma 2.5. For any f H, (Nn (f))m is a linear combination of{j}1 , where = max{m, n}.

Proof. By Definition 2.2 and the operator equality

(I 2SmA + B

)n = I +

n

l=1

n

l

2

A + B

lSlm,

we conclude that

(Nk )m = 2(N + 1)A + B k + 2A + BNn=1

nl=1

n

l

2

A + B

lSlmk.

Again by Definition 2.3, we get

(Nn (f))m =2(N + 1)

A + B

nk=1

(f, k)k +2

A + B

Nn=1

nl,k=1

bn,l,kSlmk =: M1 + M2,

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where bn,l,k =

n

l

2

A+B

l(f, k).

For any f H, by (2.2), we obtain that for any k, l Z+

,

Slmk =

mj=1

cjj, where cj =

m1,...,l1=1

(k, 1)(1 , 2) (l1 , j).

So we see that for l, k Z+, Slmk is a linear combination of m frame elements 1, 2,...,m, further, the

sum M2 is a linear combination of m elements 1, 2,...,m. Clearly, the sum M1 is a linear combination of n

elements 1, 2,...,n. Therefore, (Nn (f))m is a linear combination of{j}1 , where = max{m, n}. Lemma

2.5 is proved.

3. Approximation by (Nn ())m

We will approximate to f by (Nn (f))m. First, we estimate f Nn (f) in Lemma 3.1. Next, we

estimate Nn (f) (Nn (f))m in Lemma 3.3. Finally, we get an estimate f (Nn (f))m in Theorem 3.4.

Meanwhile, we show that the approximation error only depends on the frame bounds and the convergence rate

of the frame coefficients of f as well as the relation among frame elements.

Lemma 3.1. Let {k}1 be a frame for H with bounds A, B and Nn (f) be stated in (2.1). Denote

n(f) :=1

f

j=n+1

|(f, j)|2 12 . (3.1)

Then for any f H, we have

f Nn (f) qN+1 f +1A

1 + qN+1

f n(f) (n, N Z+), (3.2)where q = BA

B+A.

Remark 3.2. Since

{k

}1 is a frame, we see that

k=1 |(f, k)|2 < . From this and (3.1), we getn(f) 0 (n ).

Proof of Lemma 3.1. By Proposition 1.2(ii) and (2.1), we know that for any f H and N Z+, the

seriesk=1

(f, k)Nk converges and Nn (f) = nk=1

(f, k)Nk is its partial sum. Denote its remainder term byrNn (f) :=

k=n+1

(f, k)Nk .5


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So

f

Nn (f)

=

f

k=1(f, k)Nk rNn (f) f

k=1(f, k)Nk + rNn (f) .Using Proposition 1.2(ii), we get

f Nn (f) qN+1 f + rNn (f) . (3.3)

By Proposition 1.2(i), the seriesk=1

(f, k)k converges, so we can decompose rNn (f) as follows:rNn (f) =

k=n+1(f, k)

k +

k=n+1(f, k)(

Nk

k) =: u

Nn (f) + v

Nn (f). (3.4)

By (1.4) and (1.5), it follows that

k Nk = 2A+B n=N+1

(I 2SA+B

)nk

= (I 2SA+B )

N+1

2

A+B k +2

A+B

n=1

(I 2SA+B )

nk

= (I 2S

A+B)N+1k = RN+1k,

where R = I 2SA+B

. So we get

vNn (f) =

k=n+1

(f, k)RN+1k = RN+1

k=n+1

(f, k)k

.

From this and (3.4), we get

rNn (f) =

k=n+1

(f, k)k RN+1

k=n+1

(f, k)k

= (I RN+1)

k=n+1

(f, k)k. (3.5)However, we have

k=n+1

(f, k)k 2= supg=1

k=n+1

(f, k)k, g2 = supg=1

k=n+1

(f, k)(k, g) 2 .Using Cauchys inequality in l2, we get

k=n+1

(f, k)k 2

k=n+1

|(f, k)|2

supg=1

k=n+1

|(k, g)|2

.

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By Proposition 1.1 (iii), we know that {

k}1 is also a frame for H and

k=1

|(k, g)|2 1A

g 2 .

From this, we get

k=n+1

(f, k)k 2 1A

k=n+1

|(f, k)|2

.

Again by (3.1) and (3.5), we have

rNn (f) I RN+1

1

A f 2 2n(f)

12

.

By Proposition 1.1(v), we have I RN+1 1 + qN+1 (q = BAB+A

). So

rNn (f) 1A

1 + qN+1

f n(f).Finally, by (3.3), we obtain the conclusion of Lemma 3.1.

We will approximate to Nn (f) by (Nn (f))m.

Lemma 3.3. Let {k}1 be a frame for H with bounds A, B, and let Nn (f) and (Nn (f))m be stated in

(2.1) and Definition 2.3, respectively. Then for any f H, we have

Nn (f) (Nn (f))m

B f

1 +2

A + B

N+1 n N,n,m (N, n , m Z+),

where

N,n,m := max1lN

1kn Slk Slmk . (3.6)

Proof. By (1.5), we have

Nk = N1k + 2A + B k + 2A + BNl=1

N

l

2

A + B

lSlk.

and by (2.4), we have

(Nk )m = (N1k )m + 2A + B k + 2A + BNl=1

N

l

2

A + B

lSlmk (3.7)

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So we get

Nk (Nk )m = N1k (N1k )m + 2A + BNl=1

Nl 2A + Bl

(Slk Slmk).

Recursively using the above formula, we have

Nk (Nk )m = (1k (1k)m + 2A + BNj=2

jl=1

j

l

2

A + B

l(Slk Slmk). (3.8)

By Definition 2.2 and (1.5), we get

1k (

1k)m = 2

A + B2

(Sk Smk).

From this and (3.8), we have

Nk (Nk )m = 2A + BNj=1

jl=1

j

l

2

A + B

l(Slk Slmk).

So we obtain

Nk (

Nk )m

2

A + B

Nj=1

jl=1

j

l

2

A + B

l max1lN

Slk Slmk .

However,

2

A + B

Nj=1

jl=1

j

l

2

A + B

l 2

A + B

Nj=1

1 +

2

A + B

j

1 +2

A + B

N+1.

So we have

Nk (Nk )m 1 + 2A + BN+1

max1lN

Slk Slmk . (3.9)

By (2.1) and Definition 2.3, using cauchys inequality, we obtain that for any f H,

Nn (f) (Nn (f))m n

k=1|(f, k)| Nk (Nk )m

nk=1

|(f, k)|2 1

2

nk=1

Nk (Nk )m 2 12(3.10)

Combining (3.9)-(3.10) with (3.6), we get

Nn (f) (Nn (f))m

1 +2

A + B

N+1 n N,n,m

n

k=1

|(f, k)|2 1

2

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From this and (1.1), we get the conclusion of Lemma 3.3.

Since

f (Nn (f))m f Nn (f) + Nn (f) (Nn (f))m ,

by Lemma 3.1 and Lemma 3.3, we conclude immediately the following

Theorem 3.4. Let {k}1 be a frame with bounds A and B, and let

(Nn (f))m =

Nk=1

(f, k)(Nk )m (f H),where (Nk )m is stated in Definition 2.2. Then for any f H and N, n , m Z+, we have

f (Nn (f))m qN+1 f

+ 1A

1 + qN+1

f n(f) + B f 1 + 2A+BN+1 n N,n,m, (3.11)where n(f) is stated in (3.1), q =

BAB+A

, and

N,n,m = max1lN

1kn Slk Slmk (S is the frame operator).

Remark 3.5. In Lemma 2.5, we have shown that (Nn

(f))m is a linear combination of 1,..., ( =

max{m, n}). So Theorem 3.4 gives a formula approximating to any f H by a linear combination of finitely

many frame elements.

Remark 3.6. Denote

R1 := qN+1 f ,

R2 :=1

A 1 + qN+1

f n(f),R3 :=

B f

1 +

2

A + B

N+1 n N,n,m. (3.12)

Since q = BAB+A < 1, we see that R1 0 (N ). By Remark 3.2, we have R2 0 (n ). From

Lemma 2.4(ii) and (3.6), we obtain N,n,m 0 (m ), so we have R3 0 (m ).

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Therefore, for any f H and an approximation error > 0, first we choose N such that R1 < 3 , next,

we choose n such that R2

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