How to prove that the Haar system is orthonormal?
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Haar wavelets are defined as:
$$
psi_{0,0}(t) =
begin{cases}
1, text{ for } 0<t< 1/2\
-1, text{ for } 1/2<t<1 \
0, text{ otherwise }
end{cases}
$$
Where mother wavelet is$$psi_{n,k} = 2^{-n/2} psi_{0,0}(2^n t -k).$$
And for $n geq 0$, $0 leq k < 2^n$
$$
{psi_{i,n}psi_{k,l}} = delta_{i,l}delta_{k,n}$$
Show that these wavelets form an orthonormal set:
functional-analysis reference-request
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add a comment |
$begingroup$
Haar wavelets are defined as:
$$
psi_{0,0}(t) =
begin{cases}
1, text{ for } 0<t< 1/2\
-1, text{ for } 1/2<t<1 \
0, text{ otherwise }
end{cases}
$$
Where mother wavelet is$$psi_{n,k} = 2^{-n/2} psi_{0,0}(2^n t -k).$$
And for $n geq 0$, $0 leq k < 2^n$
$$
{psi_{i,n}psi_{k,l}} = delta_{i,l}delta_{k,n}$$
Show that these wavelets form an orthonormal set:
functional-analysis reference-request
$endgroup$
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Any book on wavelet theory has a proof. If there are some steps in the proof that you don't understand we will try to help you.
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– Kavi Rama Murthy
Jan 9 at 11:42
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apologies for my editing mistake previously, but I couldn't find a proof related to my query.
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– Geet Kalsulkar
Jan 9 at 12:04
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Go to statmathbc.wordpress.com, click on Catalog, click on Basic Wavelet Theory and then click on Haarlets. You can download my notes on wavelets.
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– Kavi Rama Murthy
Jan 9 at 12:24
add a comment |
$begingroup$
Haar wavelets are defined as:
$$
psi_{0,0}(t) =
begin{cases}
1, text{ for } 0<t< 1/2\
-1, text{ for } 1/2<t<1 \
0, text{ otherwise }
end{cases}
$$
Where mother wavelet is$$psi_{n,k} = 2^{-n/2} psi_{0,0}(2^n t -k).$$
And for $n geq 0$, $0 leq k < 2^n$
$$
{psi_{i,n}psi_{k,l}} = delta_{i,l}delta_{k,n}$$
Show that these wavelets form an orthonormal set:
functional-analysis reference-request
$endgroup$
Haar wavelets are defined as:
$$
psi_{0,0}(t) =
begin{cases}
1, text{ for } 0<t< 1/2\
-1, text{ for } 1/2<t<1 \
0, text{ otherwise }
end{cases}
$$
Where mother wavelet is$$psi_{n,k} = 2^{-n/2} psi_{0,0}(2^n t -k).$$
And for $n geq 0$, $0 leq k < 2^n$
$$
{psi_{i,n}psi_{k,l}} = delta_{i,l}delta_{k,n}$$
Show that these wavelets form an orthonormal set:
functional-analysis reference-request
functional-analysis reference-request
edited Jan 9 at 12:01
Geet Kalsulkar
asked Jan 9 at 11:41
Geet KalsulkarGeet Kalsulkar
33
33
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Any book on wavelet theory has a proof. If there are some steps in the proof that you don't understand we will try to help you.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:42
$begingroup$
apologies for my editing mistake previously, but I couldn't find a proof related to my query.
$endgroup$
– Geet Kalsulkar
Jan 9 at 12:04
$begingroup$
Go to statmathbc.wordpress.com, click on Catalog, click on Basic Wavelet Theory and then click on Haarlets. You can download my notes on wavelets.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 12:24
add a comment |
$begingroup$
Any book on wavelet theory has a proof. If there are some steps in the proof that you don't understand we will try to help you.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:42
$begingroup$
apologies for my editing mistake previously, but I couldn't find a proof related to my query.
$endgroup$
– Geet Kalsulkar
Jan 9 at 12:04
$begingroup$
Go to statmathbc.wordpress.com, click on Catalog, click on Basic Wavelet Theory and then click on Haarlets. You can download my notes on wavelets.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 12:24
$begingroup$
Any book on wavelet theory has a proof. If there are some steps in the proof that you don't understand we will try to help you.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:42
$begingroup$
Any book on wavelet theory has a proof. If there are some steps in the proof that you don't understand we will try to help you.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:42
$begingroup$
apologies for my editing mistake previously, but I couldn't find a proof related to my query.
$endgroup$
– Geet Kalsulkar
Jan 9 at 12:04
$begingroup$
apologies for my editing mistake previously, but I couldn't find a proof related to my query.
$endgroup$
– Geet Kalsulkar
Jan 9 at 12:04
$begingroup$
Go to statmathbc.wordpress.com, click on Catalog, click on Basic Wavelet Theory and then click on Haarlets. You can download my notes on wavelets.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 12:24
$begingroup$
Go to statmathbc.wordpress.com, click on Catalog, click on Basic Wavelet Theory and then click on Haarlets. You can download my notes on wavelets.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 12:24
add a comment |
2 Answers
2
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oldest
votes
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We need to show the following $$langle psi_{n,k_1},psi_{n,k_2}rangle=0\langle psi_{n_1,k},psi_{n_2,k}rangle=0\langle psi_{n_1,k_1},psi_{n_2,k_2}rangle=0\langle psi_{n,k},psi_{n,k}rangle=1$$when $n_1ne n_2$ and $k_1ne k_2$ and $$langle f,grangle=int_Df(x)g^*(x)dx$$
(1) for $k_1ne k_2$
$$langle psi_{n,k_1},psi_{n,k_2}rangle{=int_{Bbb R}2^{-n}psi_{0,0}(2^nx-k_1)psi_{0,0}(2^nx-k_2)dx\=2^{-2n}int_{Bbb R}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{(k_1,k_1+1)cap(k_2,k_2+1)}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{emptyset}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=0}$$
(2) for $n_1ne n_2$ withput loss of generality assume $n_1>n_2$. Therefore$$langle psi_{n_1,k},psi_{n_2,k}rangle{=int_{Bbb R}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx\=int_{({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx}$$note that for $kne 0$ $${k+1over 2^{n_1}}le {kover 2^{n_2}}$$and therefore $$({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})= emptyset$$This important result says that $langle psi_{n_1,k},psi_{n_2,k}rangle= 0$ whenever $kne 0$. In the case $k=0$ we obtain $$langle psi_{n_1,0},psi_{n_2,0}rangle{=int_{(0,{1over 2^{n_1}})cap (0,{1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1over 2^{n_1}}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)psi_{0,0}(2^{n_2-n_1}u)du\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)du=0}$$
(3)
We show that $$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx=0$$
proof
Assume $n_2>n_1$. Therefore
$$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx{={1over 2^{n_1}}int_{k_1}^{k_1+1}psi_{0,0}(u-k)psi_{0,0}(2^{n_2-n_1}u-k_2)du}\={1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du-{1over 2^{n_1}}int_{k_1+{1over 2}}^{k_1+1}psi_{0,0}(2^{n_2-n_1}u-k_2)du$$from the other side $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du=0$$ since $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du{={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}}psi_{0,0}(w-k_2)dw\={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}+k_2}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}+k_2}psi_{0,0}(w)dw\=0}$$where the last equality comes from $$int_0^1 psi_{0,0}(x)dx=0$$
(4)
$$langle psi_{n,k},psi_{n,k}rangle{=int_{Bbb R}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}dx\=1}$$which finishes our (long and exhaustive!) proof $blacksquare$
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If the supports are disjoint then of course the $L^2$ inner product is zero. If the supports overlap, the explanation for the orthogonality is as follows. A Haar wavelet has zero integral, or averages to zero. The wavelet with higher frequency (higher $n$) has support in an interval where the lower frequency wavelet is constant. So the inner product is just that constant times the spatial average of the higher frequency wavelet, i.e., zero.
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2 Answers
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2 Answers
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$begingroup$
We need to show the following $$langle psi_{n,k_1},psi_{n,k_2}rangle=0\langle psi_{n_1,k},psi_{n_2,k}rangle=0\langle psi_{n_1,k_1},psi_{n_2,k_2}rangle=0\langle psi_{n,k},psi_{n,k}rangle=1$$when $n_1ne n_2$ and $k_1ne k_2$ and $$langle f,grangle=int_Df(x)g^*(x)dx$$
(1) for $k_1ne k_2$
$$langle psi_{n,k_1},psi_{n,k_2}rangle{=int_{Bbb R}2^{-n}psi_{0,0}(2^nx-k_1)psi_{0,0}(2^nx-k_2)dx\=2^{-2n}int_{Bbb R}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{(k_1,k_1+1)cap(k_2,k_2+1)}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{emptyset}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=0}$$
(2) for $n_1ne n_2$ withput loss of generality assume $n_1>n_2$. Therefore$$langle psi_{n_1,k},psi_{n_2,k}rangle{=int_{Bbb R}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx\=int_{({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx}$$note that for $kne 0$ $${k+1over 2^{n_1}}le {kover 2^{n_2}}$$and therefore $$({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})= emptyset$$This important result says that $langle psi_{n_1,k},psi_{n_2,k}rangle= 0$ whenever $kne 0$. In the case $k=0$ we obtain $$langle psi_{n_1,0},psi_{n_2,0}rangle{=int_{(0,{1over 2^{n_1}})cap (0,{1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1over 2^{n_1}}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)psi_{0,0}(2^{n_2-n_1}u)du\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)du=0}$$
(3)
We show that $$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx=0$$
proof
Assume $n_2>n_1$. Therefore
$$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx{={1over 2^{n_1}}int_{k_1}^{k_1+1}psi_{0,0}(u-k)psi_{0,0}(2^{n_2-n_1}u-k_2)du}\={1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du-{1over 2^{n_1}}int_{k_1+{1over 2}}^{k_1+1}psi_{0,0}(2^{n_2-n_1}u-k_2)du$$from the other side $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du=0$$ since $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du{={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}}psi_{0,0}(w-k_2)dw\={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}+k_2}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}+k_2}psi_{0,0}(w)dw\=0}$$where the last equality comes from $$int_0^1 psi_{0,0}(x)dx=0$$
(4)
$$langle psi_{n,k},psi_{n,k}rangle{=int_{Bbb R}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}dx\=1}$$which finishes our (long and exhaustive!) proof $blacksquare$
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We need to show the following $$langle psi_{n,k_1},psi_{n,k_2}rangle=0\langle psi_{n_1,k},psi_{n_2,k}rangle=0\langle psi_{n_1,k_1},psi_{n_2,k_2}rangle=0\langle psi_{n,k},psi_{n,k}rangle=1$$when $n_1ne n_2$ and $k_1ne k_2$ and $$langle f,grangle=int_Df(x)g^*(x)dx$$
(1) for $k_1ne k_2$
$$langle psi_{n,k_1},psi_{n,k_2}rangle{=int_{Bbb R}2^{-n}psi_{0,0}(2^nx-k_1)psi_{0,0}(2^nx-k_2)dx\=2^{-2n}int_{Bbb R}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{(k_1,k_1+1)cap(k_2,k_2+1)}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{emptyset}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=0}$$
(2) for $n_1ne n_2$ withput loss of generality assume $n_1>n_2$. Therefore$$langle psi_{n_1,k},psi_{n_2,k}rangle{=int_{Bbb R}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx\=int_{({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx}$$note that for $kne 0$ $${k+1over 2^{n_1}}le {kover 2^{n_2}}$$and therefore $$({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})= emptyset$$This important result says that $langle psi_{n_1,k},psi_{n_2,k}rangle= 0$ whenever $kne 0$. In the case $k=0$ we obtain $$langle psi_{n_1,0},psi_{n_2,0}rangle{=int_{(0,{1over 2^{n_1}})cap (0,{1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1over 2^{n_1}}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)psi_{0,0}(2^{n_2-n_1}u)du\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)du=0}$$
(3)
We show that $$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx=0$$
proof
Assume $n_2>n_1$. Therefore
$$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx{={1over 2^{n_1}}int_{k_1}^{k_1+1}psi_{0,0}(u-k)psi_{0,0}(2^{n_2-n_1}u-k_2)du}\={1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du-{1over 2^{n_1}}int_{k_1+{1over 2}}^{k_1+1}psi_{0,0}(2^{n_2-n_1}u-k_2)du$$from the other side $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du=0$$ since $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du{={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}}psi_{0,0}(w-k_2)dw\={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}+k_2}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}+k_2}psi_{0,0}(w)dw\=0}$$where the last equality comes from $$int_0^1 psi_{0,0}(x)dx=0$$
(4)
$$langle psi_{n,k},psi_{n,k}rangle{=int_{Bbb R}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}dx\=1}$$which finishes our (long and exhaustive!) proof $blacksquare$
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We need to show the following $$langle psi_{n,k_1},psi_{n,k_2}rangle=0\langle psi_{n_1,k},psi_{n_2,k}rangle=0\langle psi_{n_1,k_1},psi_{n_2,k_2}rangle=0\langle psi_{n,k},psi_{n,k}rangle=1$$when $n_1ne n_2$ and $k_1ne k_2$ and $$langle f,grangle=int_Df(x)g^*(x)dx$$
(1) for $k_1ne k_2$
$$langle psi_{n,k_1},psi_{n,k_2}rangle{=int_{Bbb R}2^{-n}psi_{0,0}(2^nx-k_1)psi_{0,0}(2^nx-k_2)dx\=2^{-2n}int_{Bbb R}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{(k_1,k_1+1)cap(k_2,k_2+1)}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{emptyset}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=0}$$
(2) for $n_1ne n_2$ withput loss of generality assume $n_1>n_2$. Therefore$$langle psi_{n_1,k},psi_{n_2,k}rangle{=int_{Bbb R}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx\=int_{({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx}$$note that for $kne 0$ $${k+1over 2^{n_1}}le {kover 2^{n_2}}$$and therefore $$({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})= emptyset$$This important result says that $langle psi_{n_1,k},psi_{n_2,k}rangle= 0$ whenever $kne 0$. In the case $k=0$ we obtain $$langle psi_{n_1,0},psi_{n_2,0}rangle{=int_{(0,{1over 2^{n_1}})cap (0,{1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1over 2^{n_1}}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)psi_{0,0}(2^{n_2-n_1}u)du\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)du=0}$$
(3)
We show that $$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx=0$$
proof
Assume $n_2>n_1$. Therefore
$$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx{={1over 2^{n_1}}int_{k_1}^{k_1+1}psi_{0,0}(u-k)psi_{0,0}(2^{n_2-n_1}u-k_2)du}\={1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du-{1over 2^{n_1}}int_{k_1+{1over 2}}^{k_1+1}psi_{0,0}(2^{n_2-n_1}u-k_2)du$$from the other side $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du=0$$ since $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du{={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}}psi_{0,0}(w-k_2)dw\={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}+k_2}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}+k_2}psi_{0,0}(w)dw\=0}$$where the last equality comes from $$int_0^1 psi_{0,0}(x)dx=0$$
(4)
$$langle psi_{n,k},psi_{n,k}rangle{=int_{Bbb R}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}dx\=1}$$which finishes our (long and exhaustive!) proof $blacksquare$
$endgroup$
We need to show the following $$langle psi_{n,k_1},psi_{n,k_2}rangle=0\langle psi_{n_1,k},psi_{n_2,k}rangle=0\langle psi_{n_1,k_1},psi_{n_2,k_2}rangle=0\langle psi_{n,k},psi_{n,k}rangle=1$$when $n_1ne n_2$ and $k_1ne k_2$ and $$langle f,grangle=int_Df(x)g^*(x)dx$$
(1) for $k_1ne k_2$
$$langle psi_{n,k_1},psi_{n,k_2}rangle{=int_{Bbb R}2^{-n}psi_{0,0}(2^nx-k_1)psi_{0,0}(2^nx-k_2)dx\=2^{-2n}int_{Bbb R}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{(k_1,k_1+1)cap(k_2,k_2+1)}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=2^{-2n}int_{emptyset}psi_{0,0}(u-k_1)psi_{0,0}(u-k_2)du\=0}$$
(2) for $n_1ne n_2$ withput loss of generality assume $n_1>n_2$. Therefore$$langle psi_{n_1,k},psi_{n_2,k}rangle{=int_{Bbb R}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx\=int_{({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x-k)psi_{0,0}(2^{n_2}x-k)dx}$$note that for $kne 0$ $${k+1over 2^{n_1}}le {kover 2^{n_2}}$$and therefore $$({kover 2^{n_1}},{k+1over 2^{n_1}})cap ({kover 2^{n_2}},{k+1over 2^{n_2}})= emptyset$$This important result says that $langle psi_{n_1,k},psi_{n_2,k}rangle= 0$ whenever $kne 0$. In the case $k=0$ we obtain $$langle psi_{n_1,0},psi_{n_2,0}rangle{=int_{(0,{1over 2^{n_1}})cap (0,{1over 2^{n_2}})}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1over 2^{n_1}}2^{-{n_1+n_2over 2}}psi_{0,0}(2^{n_1}x)psi_{0,0}(2^{n_2}x)dx\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)psi_{0,0}(2^{n_2-n_1}u)du\=int_0^{1}2^{-{3n_1+n_2over 2}}psi_{0,0}(u)du=0}$$
(3)
We show that $$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx=0$$
proof
Assume $n_2>n_1$. Therefore
$$int_{Bbb R}psi_{0,0}(2^{n_1}x-k_1)psi_{0,0}(2^{n_2}x-k_2)dx{={1over 2^{n_1}}int_{k_1}^{k_1+1}psi_{0,0}(u-k)psi_{0,0}(2^{n_2-n_1}u-k_2)du}\={1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du-{1over 2^{n_1}}int_{k_1+{1over 2}}^{k_1+1}psi_{0,0}(2^{n_2-n_1}u-k_2)du$$from the other side $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du=0$$ since $${1over 2^{n_1}}int_{k_1}^{k_1+{1over 2}}psi_{0,0}(2^{n_2-n_1}u-k_2)du{={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}}psi_{0,0}(w-k_2)dw\={1over 2^{n_2}}int_{k_1cdot 2^{n_2-n_1}+k_2}^{left(k_1+{1over 2}right)cdot 2^{n_2-n_1}+k_2}psi_{0,0}(w)dw\=0}$$where the last equality comes from $$int_0^1 psi_{0,0}(x)dx=0$$
(4)
$$langle psi_{n,k},psi_{n,k}rangle{=int_{Bbb R}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}psi_{0,0}^2(2^nx-k)dx\=int_{kover 2^n}^{k+1over 2^n}2^{-n}dx\=1}$$which finishes our (long and exhaustive!) proof $blacksquare$
answered Jan 9 at 13:31
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
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$begingroup$
If the supports are disjoint then of course the $L^2$ inner product is zero. If the supports overlap, the explanation for the orthogonality is as follows. A Haar wavelet has zero integral, or averages to zero. The wavelet with higher frequency (higher $n$) has support in an interval where the lower frequency wavelet is constant. So the inner product is just that constant times the spatial average of the higher frequency wavelet, i.e., zero.
$endgroup$
add a comment |
$begingroup$
If the supports are disjoint then of course the $L^2$ inner product is zero. If the supports overlap, the explanation for the orthogonality is as follows. A Haar wavelet has zero integral, or averages to zero. The wavelet with higher frequency (higher $n$) has support in an interval where the lower frequency wavelet is constant. So the inner product is just that constant times the spatial average of the higher frequency wavelet, i.e., zero.
$endgroup$
add a comment |
$begingroup$
If the supports are disjoint then of course the $L^2$ inner product is zero. If the supports overlap, the explanation for the orthogonality is as follows. A Haar wavelet has zero integral, or averages to zero. The wavelet with higher frequency (higher $n$) has support in an interval where the lower frequency wavelet is constant. So the inner product is just that constant times the spatial average of the higher frequency wavelet, i.e., zero.
$endgroup$
If the supports are disjoint then of course the $L^2$ inner product is zero. If the supports overlap, the explanation for the orthogonality is as follows. A Haar wavelet has zero integral, or averages to zero. The wavelet with higher frequency (higher $n$) has support in an interval where the lower frequency wavelet is constant. So the inner product is just that constant times the spatial average of the higher frequency wavelet, i.e., zero.
answered Jan 9 at 12:21
Abdelmalek AbdesselamAbdelmalek Abdesselam
869312
869312
add a comment |
add a comment |
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$begingroup$
Any book on wavelet theory has a proof. If there are some steps in the proof that you don't understand we will try to help you.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:42
$begingroup$
apologies for my editing mistake previously, but I couldn't find a proof related to my query.
$endgroup$
– Geet Kalsulkar
Jan 9 at 12:04
$begingroup$
Go to statmathbc.wordpress.com, click on Catalog, click on Basic Wavelet Theory and then click on Haarlets. You can download my notes on wavelets.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 12:24