L^{p} NON-INVERTIBILITY OF FRAME OPERATORS

TERENCE TAO

Abstract. We give an example of an operator of the form

T f = ∑_{j,k}⟨f, Ψ_{j,k}⟩Ψ_{j,k},

where Ψ is a pseudo-wavelet, which is bounded and invertible on L^{2} but not on L^{p} for any specified 1 < p < 2.

1. Introduction

Fix 1 < p < 2. Let ψ be a smooth compactly supported mother wavelet (the exact choice of which is unimportant), and let

ψ_{j,k}(x) = 2^{−j/2}ψ(2^{−j}x − k)

be the associated orthonormal basis of L^{2}. Define

Ψ = ψ_{0,0} − αψ_{−1,0}

where 0 < α < 1 is given by

α = 2^{\frac{1}{2} − \frac{1}{p}} .

Note that

∥ψ_{j,k}∥_{p} = α^{−j}∥ψ_{0,0}∥_{p}.

Also, if we define

Ψ_{j,k}(x) = 2^{−j/2}Ψ(2^{−j}x − k)

then we have

Ψ_{j,k} = ψ_{j,k} − αψ_{j−1,2k}.

Clearly

(1 − α)∥f∥_{2} ≤ (∑_{j,k} |⟨f, Ψ_{j,k}⟩|^{2})^{1/2} ≤ (1 + α)∥f∥_{2}

so the Ψ_{j,k} form a frame.

Let N be a large integer and define

f = ∑_{i=0}^{N} α^{i}ψ_{i,0}.

From the Littlewood-Paley theory for ψ we have

∥f∥_{p} ∼ ∥(∑_{i=0}^{N} α^{2i}|ψ_{i,0}|^{2})^{1/2}∥_{p} ∼ N^{1/p}

as an easy calculation shows.

1991 Mathematics Subject Classification. 42B.

On the other hand, we have

⟨f, Ψ_{0,0}⟩ = 1
⟨f, Ψ_{N+1,0}⟩ = −α^{N+1}

and that ⟨f, Ψ_{j,k}⟩ = 0 for all other Ψ_{j,k}. Thus

∑_{j,k}⟨f, Ψ_{j,k}⟩Ψ_{j,k} = Ψ_{0,0} − α^{N+1}Ψ_{N+1,0}.

Thus

∥ ∑_{j,k}⟨f, Ψ_{j,k}⟩Ψ_{j,k}∥_{p} ≲ 1.

Thus we see that the operator

f ↦ ∑_{j,k}⟨f, Ψ_{j,k}⟩Ψ_{j,k}

is not invertible on L^{p}, even though it is invertible on L^{2}. (In particular, the inverse of this operator is not a Calderón-Zygmund operator).

The author thanks Amos Ron for discussions which originated this problem, and for pointing out some errors in the original version of this note.

Department of Mathematics, UCLA, Los Angeles, CA 90024

   E-mail address: tao@math.ucla.edu