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Gelfand n-widths

Module by: Mark Davenport, Ronald DeVore, Chris Rozell, Michael Wakin

Continuing from last time, we have a signal xR N xR N , an n×Nn×N measurement matrix ΦΦ, and y=ΦxR n y=ΦxR n is the information we draw from xx.

We consider decoders ΔΔ mapping R n R N R n R N . We have been discussing whether there exists a decoder with certain properties. So for this discussion (about information preservation), we can just think about optimal decoding.

While the previous result on sparse signal recovery is interesting, it is not very robust. What happens if our signal does not have support kk? Can we still obtain meaningful results? The answer is yes. To do so we introduce more general input signal classes KXKX that allows fully supported signals. For example, we will consider the signal class defined by the unit ball

K = U ( p N ) = { x : x X 1 } . K=U( p N )={x: x X 1}. (1)
Given an encoder/decoder pair ( Φ , Δ ) (Φ,Δ), the worst case error on a set KK for that pair will be given by
E ( K , Φ , Δ ) X = sup x K x Δ ( Φ x ) X . E ( K , Φ , Δ ) X = sup x K x Δ ( Φ x ) X .(2)
Finally, using min-max principles we will define the minimum error over all encoder/decoder pairs for a signal class and for a fixed number of measurements nn to be
E n ( K ) X = inf ( Φ , Δ ) : Φ is n×N E ( K , Φ , Δ ) X . E n ( K ) X = inf ( Φ , Δ ) : Φ is n×N E ( K , Φ , Δ ) X . (3)
This measure E n (K) X E n (K) X is the best we could do while measuring distortion on the topology of XX, using nn linear measurements, and using arbitrary decoding.

We will see that these questions are actually equivalent to a classical “nn-width” problem. nn-widths have seen a great deal of work over the years by a variety of mathematicians: Kolmogorov, Tikhomirov, Kashin, Gluskin, etc. There are many different flavors of nn-widths, but we will study the Gelfand nn-width (the least intuitive of the widths).

definition 1: Gelfand n-width

Let KXKX be compact. Given nn, the Gelfand width (also called the dual width) is given by

d n (K) X := inf Y : codim (Y) = n sup { x X : xKY} . d n (K) X := inf Y : codim (Y) = n sup{ x X :xKY}.(4)
where by codimension (Y)=n codimension (Y)=n we mean that YY has dimension dim (X)n dim (X)n.

In other words, we are looking for the subspace YY that slices through the set KK so that the norms of the projected signals are as small as possible. We can now state the following theorem about n-widths:

theorem 1

Provided that KK has the properties (1) K=-KK=-K and (2) K+K=CKK+K=CK, then

d n (K) X E n (K) X Cd n (K) X d n (K) X E n (K) X Cd n (K) X (5)
where CC is the same constant listed in property (2).1

Proof

We start with the left-hand inequality. We want to take any encoder/decoder pair and use that to construct a YY. So let Φ,ΔΦ,Δ be an encoder/decoder. Then simply let Y=N(Φ)Y=N(Φ). Now consider an ηKYηKY and note that Φ(η)=0Φ(η)=0 since ηYηY. Let z=Δ(0)z=Δ(0) be the decoding of 0 (practically speaking, zz should be zero itself, but we avoid that assumption in this proof). Then

η X max(η-z X ,η+z X )=max(η-ΔΦ(η) X ,-η-ΔΦ(η) X ) sup xK x - Δ Φ (x) X η X max(η-z X ,η+z X )=max(η-ΔΦ(η) X ,-η-ΔΦ(η) X ) sup xK x - Δ Φ (x) X (6)
where we first employ the triangle inequality, then the fact that multiplying by -1-1 does not change the norm, then the fact that K=-KK=-K. So then
sup ηKY η X sup xK x - Δ Φ ( x ) X . sup ηKY η X sup xK x - Δ Φ ( x ) X .(7)
Taking the infimum over all Φ,ΔΦ,Δ, it follows that
d n (K) X E n (K) X . d n (K) X E n (K) X .(8)
Since ΦΦ is n×Nn×N, then dim (N(Φ))N-n dim (N(Φ))N-n.

Now we prove the right-hand inequality. Assume we have a good YY. Suppose YY has codimension N-nN-n. Then Y Y (the orthogonal complement of YY in R N R N ) has dimension nn. Let v 1 ,v 2 ,,v n R N v 1 ,v 2 ,,v n R N be a basis for Y Y . Let ΦΦ be the n×Nn×N matrix obtained by stacking the rows v 1 ,v 2 ,,v n v 1 ,v 2 ,,v n . Then N(Φ)=YN(Φ)=Y. Define Δ(y)=Δ(y)= any element of KF(y)KF(y) if there is one (otherwise let Δ(y)Δ(y) be anything in F(y)F(y)). Now look at the performance x-ΔΦ(x) X x-ΔΦ(x) X for some xKxK. Both xx and ΔΦ(x)=:x ' ΔΦ(x)=:x ' are elements of KK, so x-x ' x-x ' is in N(Φ)N(Φ) and in CKCK. Therefore x-x ' CKN(Φ)x-x ' CKN(Φ). Thus,

x - x ' C X sup z Y K z X , x - x ' C X sup z Y K z X ,(9)
and so for any xKxK,
x-ΔΦ(x) X Csup zYK z X .x-ΔΦ(x) X Csup zYK z X .(10)
Taking the infimum over all YY, we get that E n (K) X C d n (K) X E n (K) X C d n (K) X .

From the proof of this theorem, we see that there is a matching between the matrices ΦΦ and the spaces YY (via the nullspace of ΦΦ).

An important result is that d n (U( q N ) ) p N d n (U( q N ) ) p N is known for all p,qp,q except p=1,q=p=1,q=. A precise statement of these widths can be found in the book (Reference). A particularly important case is

C 1 log ( N / n ) n d n ( U ( 1 N ) ) 2 N C 2 log ( N / n ) n C 1 log ( N / n ) n d n ( U ( 1 N ) ) 2 N C 2 log ( N / n ) n (11)
for N>2nN>2n. This result was first proved by Kashin with a worse logarithm in the upper inequality and later brought to the present form by Gluskin. This result solves several important problems in functional analysis and approximation.

Footnotes

  1. Clarifying notation: CK={Cx:xK}CK={Cx:xK} and K+K={x1+x2:x1,x2K}K+K={x1+x2:x1,x2K}.

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