Let x be a signal to be sparsely decomposed over a redundant dictionary A,
i.e., a sparse coefficient vector s has to be found such that x=As. It is known
that this problem is inherently unstable against noise, and to overcome this
instability, the authors of [Stable Recovery; Donoho et.al., 2006] have
proposed to use an "approximate" decomposition, that is, a decomposition
satisfying ||x - A s|| < \delta, rather than satisfying the exact equality x =
As. Then, they have shown that if there is a decomposition with ||s||_0 <
(1+M^{-1})/2, where M denotes the coherence of the dictionary, this
decomposition would be stable against noise. On the other hand, it is known
that a sparse decomposition with ||s||_0 < spark(A)/2 is unique. In other
words, although a decomposition with ||s||_0 < spark(A)/2 is unique, its
stability against noise has been proved only for highly more restrictive
decompositions satisfying ||s||_0 < (1+M^{-1})/2, because usually (1+M^{-1})/2
<< spark(A)/2.
This limitation maybe had not been very important before, because ||s||_0 <
(1+M^{-1})/2 is also the bound which guaranties that the sparse decomposition
can be found via minimizing the L1 norm, a classic approach for sparse
decomposition. However, with the availability of new algorithms for sparse
decomposition, namely SL0 and Robust-SL0, it would be important to know whether
or not unique sparse decompositions with (1+M^{-1})/2 < ||s||_0 < spark(A)/2
are stable. In this paper, we show that such decompositions are indeed stable.
In other words, we extend the stability bound from ||s||_0 < (1+M^{-1})/2 to
the whole uniqueness range ||s||_0 < spark(A)/2. In summary, we show that "all
unique sparse decompositions are stably recoverable". Moreover, we see that
sparser decompositions are "more stable".Comment: Accepted in IEEE Trans on SP on 4 May 2010. (c) 2010 IEEE. Personal
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