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Introduction to Finite Frame Theory - Frame Research Center

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<strong>Introduction</strong> <strong>to</strong> <strong>Finite</strong> <strong>Frame</strong> <strong>Theory</strong> 29<br />

5.3 <strong>Frame</strong> Algorithms<br />

Let (ϕ i ) M i=1 be a frame for H N with frame opera<strong>to</strong>r S, and assume we are given the<br />

image of a signal x ∈ H N under the analysis opera<strong>to</strong>r, i.e., the sequence (〈x,ϕ i 〉) M i=1<br />

in l M 2 . Theorem 8 already provided us with the reconstruction formula<br />

x =<br />

M<br />

∑<br />

i=1<br />

〈x,ϕ i 〉S −1 ϕ i<br />

by using the canonical dual frame. Since inversion is typically not only computationally<br />

expensive, but also numerically instable, this formula might not be utilizable<br />

in practice.<br />

To resolve this problem, we will next discuss three iterative methods <strong>to</strong> derive<br />

a converging sequence of approximations of x from knowledge of (〈x,ϕ i 〉) M i=1 . The<br />

first on our list is the so-called frame algorithm.<br />

Proposition 18 (<strong>Frame</strong> Algorithm). Let (ϕ i ) M i=1 be a frame for H N with frame<br />

bounds A,B and frame opera<strong>to</strong>r S. Given a signal x ∈ H N , define a sequence<br />

(y j ) ∞ j=0 in H N by<br />

y 0 = 0, y j = y j−1 + 2<br />

A + B S(x − y j−1) for all j ≥ 1.<br />

Then (y j ) ∞ j=0 converges <strong>to</strong> x in H N and the rate of convergence is<br />

‖x − y j ‖ ≤<br />

Proof. First, for all x ∈ H N , we have<br />

〈(<br />

Id − 2 ) 〉<br />

A + B S x,x = ‖x‖ 2 − 2<br />

A + B<br />

Similarly, we obtain<br />

which yields<br />

By the definition of y j , for any j ≥ 0,<br />

( ) B − A j<br />

‖x‖, j ≥ 0.<br />

B + A<br />

M<br />

∑<br />

i=1<br />

|〈x,ϕ i 〉| 2 ≤ ‖x‖ 2 − 2A<br />

A + B ‖x‖2 = B − A<br />

A + B ‖x‖2 .<br />

− B − A 〈(<br />

B + A ‖x‖2 ≤ Id − 2 ) 〉<br />

A + B S x,x ,<br />

∥<br />

∥Id − 2<br />

A + B S ∥ ∥∥ ≤<br />

B − A<br />

A + B . (3)<br />

x − y j = x − y j−1 − 2<br />

(<br />

A + B S(x − y j−1) = Id − 2 )<br />

A + B S (x − y j−1 ).<br />

Iterating this calculation, we derive

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