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\begin{document}

\title[$L^p$ non-invertibility]{$L^p$ non-invertibility of frame operators}

\author{Terence Tao}
\address{Department of Mathematics, UCLA, Los Angeles, CA 90024}
\email{tao@@math.ucla.edu}

\subjclass{42B}
\begin{abstract}
We give an example of an operator of the form 
$$ T f = \sum_{j,k} \langle f, \Psi_{j,k} \rangle \Psi_{j,k},$$
where $\Psi$ is a pseudo-wavelet,
which is bounded and invertible on $L^2$ but not on $L^p$ for any specified $1 < p < 2$.  
\end{abstract}

\maketitle

\section{Introduction}

Fix $1 < p < 2$.  Let $\psi$ be a smooth compactly supported mother wavelet (the exact choice of which is unimportant), and let
$$ \psi_{j,k}(x) = 2^{-j/2} \psi(2^{-j}x-k)$$
be the associated orthonormal basis of $L^2$.  Define
$$ \Psi = \psi_{0,0} - \alpha \psi_{-1,0}$$
where $0 < \alpha < 1$ is given by
$$ \alpha = 2^{\frac{1}{2} - \frac{1}{p}}.$$
Note that
$$ \| \psi_{j,k} \|_p = \alpha^{-j} \| \psi_{0,0} \|_p.$$
Also, if we define
$$ \Psi_{j,k}(x) = 2^{-j/2} \Psi(2^{-j}x-k)$$
then we have
$$ \Psi_{j,k} = \psi_{j,k} - \alpha \psi_{j-1,2k}.$$
Clearly
$$ (1-\alpha) \|f\|_2 \leq (\sum_{j,k} |\langle f, \Psi_{j,k} \rangle|^2)^{1/2} \leq (1+\alpha) \|f\|_2$$
so the $\Psi_{j,k}$ form a frame.

Let $N$ be a large integer and define
$$ f = \sum_{i=0}^N \alpha^i \psi_{i,0}.$$
From the Littlewood-Paley theory for $\psi$ we have
$$ \|f\|_p \sim \| (\sum_{i=0}^N \alpha^{2i} |\psi_{i,0}|^2)^{1/2} \|_p
\sim N^{1/p}$$
as an easy calculation shows.

On the other hand, we have
$$ \langle f, \Psi_{0,0} \rangle = 1$$
$$ \langle f, \Psi_{N+1,0} \rangle = -\alpha^{N+1}$$
and that $\langle f, \Psi_{j,k} \rangle = 0$ for all other $\Psi_{j,k}$.  Thus
$$ \sum_{j,k} \langle f, \Psi_{j,k} \rangle \Psi_{j,k} = \Psi_{0,0} - \alpha^{N+1} \Psi_{N+1,0}.$$
Thus
$$ \| \sum_{j,k} \langle f, \Psi_{j,k} \rangle \Psi_{j,k} \|_p \lesssim 1.$$
Thus we see that the operator
$$ f \mapsto \sum_{j,k} \langle f, \Psi_{j,k} \rangle \Psi_{j,k}$$
is not invertible on $L^p$, even though it is invertible on $L^2$.
(In particular, the inverse of this operator is not a Calder\'on-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.

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