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> Volume 3, Issue 11
Question: Mistake in Daubechies' paper ?

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steve@pitacat.math.cwru.edu (Steven H. Izen) Guest

Posted: Mon Dec 02, 2002 1:12 pm Subject: Question: Mistake in Daubechies' paper ?




Question: Mistake in Daubechies' paper ?
Q. Is there an error in propostion 2.11 of Daubchies' paper,``The
Wavelet Transform, Timefrequency Localization and Signal Analysis,"
IEEE Trans. Inf. Theory, 36(5), 1990?
I have prepared a one page LaTeX document (below) explaining where I think
the error might be. (I believe it comes down to the CauchySchwarz
inequality being applied in the wrong direction).
Thanks for any help.
Steve
Steve Izen, Associate Prof. of Mathematics, Case Western Reserve University
Best Address: shi@po.cwru.edu Next best address: steve@pitacat.math.cwru.edu
Phones: Analog (Voice): (216)3682891 <> Digital (Fax): (216)3685163

documentstyle{article}
egin{document}
In Daubechies' paper ``The Wavelet Transform, Timefrequency
Localization and Signal Analysis", in propositions 2.11 and 2.12 (pages
990992) conditions are given for frames in Sobolev spaces. The
construction is a generalization of the corresponding construction
for $L^2$. (See Section 3.3.2 of Daubechies' ``Ten Lectures on
Wavelets.")
However, either there is an error in the proof, or I am missing
something. From the last line on page 990,
egin{eqnarray*}
langle f,{T}f
angle &=&frac{2pi}{q_0}int
dxleft[(1+x^2)^{s/2}lefthat
f(x)
ight
ight]left[(1+x^2)^{s/2}lefthat
f(x)
ight
ight]cdot\
&&cdot
left(sum_mlefthat g(x+mp_0)
ight^2
ight) + r
end{eqnarray*}
oindent and the bound for $r$ computed on the next page, $rleq
frac{2pi}{q_0}Rf_sf_{s}$, the lower bound
$$
frac{2pi}{q_0}left(inf_xleft(sum_mlefthat
g(x+mp_0)
ight^2
ight)  R
ight)f_sf_{s} leq langle
f,{T}f
angle
$$
is deduced.
It appears to me that the first term in the left hand side of the
inequality is obtained by writing
egin{eqnarray*}
lefteqn{int
dxleft[(1+x^2)^{s/2}lefthat
f(x)
ight
ight]left[(1+x^2)^{s/2}lefthat f(x)
ight
ight]
left(sum_mlefthat g(x+mp_0)
ight^2
ight)}\
&geq&inf_xleft(sum_mlefthat g(x+mp_0)
ight^2
ight)int
dxleft[(1+x^2)^{s/2}lefthat
f(x)
ight
ight]left[(1+x^2)^{s/2}lefthat
f(x)
ight
ight]. \
end{eqnarray*}
If one had the inequality
$$
int
dxleft[(1+x^2)^{s/2}lefthat
f(x)
ight
ight]left[(1+x^2)^{s/2}lefthat
f(x)
ight
ight]geq f_sf_{s},
$$
then the result would follow, but CauchySchwarz gives the inequality
in the opposite direction.
Thus, my question is, have I missed something? Is there anyway to fix
up the argument to obtain a legitimate lower bound? As a side
comment, for $L^2$, the argument goes through with no trouble since
the last step above isn't needed.
end{document} 





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