A step in the equality $ widehat(T^*u)=|det T|^-1((T^t)^-1))^*hatu $

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Suppose $ uin S^'(mathbbR^n) $, where $ S^'(mathbbR^n) $ is the dual space of the Schwartz-class functions $ S $; $ T^*u $ is the pullback via
$ T $, i.e. if $ T: mathbbR^ntomathbbR^n $ is a linear bijection and $
uin S^'(mathbbR^n)cap C(mathbbR^n) $, we can define its pullback under $ T $ by
$$
T^*u=ucirc T.
$$

Note that a change of variables gives

$$
beginalign (T^*u)(phi)&=int u(Tx)phi(x)dx\
&= int u(y)|det T^-1|phi (T^-1y)dy\
&= u(|det T^-1|phi(T^-1)),
endalign$$
so, for general $ uin S^'(mathbbR^n) $, we define the pullback using the left and right sides of this equality.

With the above notation, prove that:

$$
widehat(T^*u)=|det T|^-1((T^t)^-1))^*hatu.
$$

The Fourier transform is defined to be $$ hatf(xi)=int_mathbbR^ne^-i<x, xi>f(x)dx $$
for $ f(x)in L^1(mathbbR^n) $.

---

Here is what I have:
$$
beginalign
(widehatT^*u)(phi(x)) &= (T^*u)(hatphi)\
&=int_mathbbR^nT^*(u(x))hatphi(x)dx\
&= int_mathbbR^nwidehatT^*[u(x)]phi(x)dx\
&= int_mathbbR^nwidehatu[T(x)]phi(x)dx\
&= int_mathbbR^nleft[int_mathbbR^ne^-i<s, xi>u[T(s)]dsright]phi(x)dx\
endalign
$$

How to continue? I am wondering that where does the transpose come from?

functional-analysis analysis fourier-analysis

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edited Jul 18 at 14:46

asked Jul 18 at 14:24

Yuchen

751114

add a comment | 

up vote
1
down vote

favorite

Suppose $ uin S^'(mathbbR^n) $, where $ S^'(mathbbR^n) $ is the dual space of the Schwartz-class functions $ S $; $ T^*u $ is the pullback via
$ T $, i.e. if $ T: mathbbR^ntomathbbR^n $ is a linear bijection and $
uin S^'(mathbbR^n)cap C(mathbbR^n) $, we can define its pullback under $ T $ by
$$
T^*u=ucirc T.
$$

Note that a change of variables gives

$$
beginalign (T^*u)(phi)&=int u(Tx)phi(x)dx\
&= int u(y)|det T^-1|phi (T^-1y)dy\
&= u(|det T^-1|phi(T^-1)),
endalign$$
so, for general $ uin S^'(mathbbR^n) $, we define the pullback using the left and right sides of this equality.

With the above notation, prove that:

$$
widehat(T^*u)=|det T|^-1((T^t)^-1))^*hatu.
$$

The Fourier transform is defined to be $$ hatf(xi)=int_mathbbR^ne^-i<x, xi>f(x)dx $$
for $ f(x)in L^1(mathbbR^n) $.

---

Here is what I have:
$$
beginalign
(widehatT^*u)(phi(x)) &= (T^*u)(hatphi)\
&=int_mathbbR^nT^*(u(x))hatphi(x)dx\
&= int_mathbbR^nwidehatT^*[u(x)]phi(x)dx\
&= int_mathbbR^nwidehatu[T(x)]phi(x)dx\
&= int_mathbbR^nleft[int_mathbbR^ne^-i<s, xi>u[T(s)]dsright]phi(x)dx\
endalign
$$

How to continue? I am wondering that where does the transpose come from?

functional-analysis analysis fourier-analysis

share|cite|improve this question

edited Jul 18 at 14:46

asked Jul 18 at 14:24

Yuchen

751114

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Suppose $ uin S^'(mathbbR^n) $, where $ S^'(mathbbR^n) $ is the dual space of the Schwartz-class functions $ S $; $ T^*u $ is the pullback via
$ T $, i.e. if $ T: mathbbR^ntomathbbR^n $ is a linear bijection and $
uin S^'(mathbbR^n)cap C(mathbbR^n) $, we can define its pullback under $ T $ by
$$
T^*u=ucirc T.
$$

Note that a change of variables gives

$$
beginalign (T^*u)(phi)&=int u(Tx)phi(x)dx\
&= int u(y)|det T^-1|phi (T^-1y)dy\
&= u(|det T^-1|phi(T^-1)),
endalign$$
so, for general $ uin S^'(mathbbR^n) $, we define the pullback using the left and right sides of this equality.

With the above notation, prove that:

$$
widehat(T^*u)=|det T|^-1((T^t)^-1))^*hatu.
$$

The Fourier transform is defined to be $$ hatf(xi)=int_mathbbR^ne^-i<x, xi>f(x)dx $$
for $ f(x)in L^1(mathbbR^n) $.

---

Here is what I have:
$$
beginalign
(widehatT^*u)(phi(x)) &= (T^*u)(hatphi)\
&=int_mathbbR^nT^*(u(x))hatphi(x)dx\
&= int_mathbbR^nwidehatT^*[u(x)]phi(x)dx\
&= int_mathbbR^nwidehatu[T(x)]phi(x)dx\
&= int_mathbbR^nleft[int_mathbbR^ne^-i<s, xi>u[T(s)]dsright]phi(x)dx\
endalign
$$

How to continue? I am wondering that where does the transpose come from?

functional-analysis analysis fourier-analysis

share|cite|improve this question

edited Jul 18 at 14:46

asked Jul 18 at 14:24

Yuchen

751114

Suppose $ uin S^'(mathbbR^n) $, where $ S^'(mathbbR^n) $ is the dual space of the Schwartz-class functions $ S $; $ T^*u $ is the pullback via
$ T $, i.e. if $ T: mathbbR^ntomathbbR^n $ is a linear bijection and $
uin S^'(mathbbR^n)cap C(mathbbR^n) $, we can define its pullback under $ T $ by
$$
T^*u=ucirc T.
$$

Note that a change of variables gives

$$
beginalign (T^*u)(phi)&=int u(Tx)phi(x)dx\
&= int u(y)|det T^-1|phi (T^-1y)dy\
&= u(|det T^-1|phi(T^-1)),
endalign$$
so, for general $ uin S^'(mathbbR^n) $, we define the pullback using the left and right sides of this equality.

With the above notation, prove that:

$$
widehat(T^*u)=|det T|^-1((T^t)^-1))^*hatu.
$$

The Fourier transform is defined to be $$ hatf(xi)=int_mathbbR^ne^-i<x, xi>f(x)dx $$
for $ f(x)in L^1(mathbbR^n) $.

---

Here is what I have:
$$
beginalign
(widehatT^*u)(phi(x)) &= (T^*u)(hatphi)\
&=int_mathbbR^nT^*(u(x))hatphi(x)dx\
&= int_mathbbR^nwidehatT^*[u(x)]phi(x)dx\
&= int_mathbbR^nwidehatu[T(x)]phi(x)dx\
&= int_mathbbR^nleft[int_mathbbR^ne^-i<s, xi>u[T(s)]dsright]phi(x)dx\
endalign
$$

How to continue? I am wondering that where does the transpose come from?

functional-analysis analysis fourier-analysis

share|cite|improve this question

edited Jul 18 at 14:46

asked Jul 18 at 14:24

Yuchen

751114

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 14:46

edited Jul 18 at 14:46

edited Jul 18 at 14:46

asked Jul 18 at 14:24

Yuchen

751114

asked Jul 18 at 14:24

Yuchen

751114

asked Jul 18 at 14:24

Yuchen

751114

751114

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

You were off to a good start. For any $phi in cal S(mathbb R^n)$ the definitions provide
$$widehatT^*u(phi) = int_mathbb R^n T^*u(x) hat phi(x) , dx = int_mathbb R^n u(Tx) hat phi(x), dx.$$

Now change variables: with $y = Tx$ you get $dy = |det T| dx$ so that
$$ int_mathbb R^n u(Tx) hat phi(x), dx = |det T|^-1 int_mathbb R^n u(y) hat phi (T^-1y) , dy.$$

Time to investigate the Fourier transform. Observe
$$ hat phi (T^-1 y) = int_mathbb R^n e^-i langle z,T^-1y rangle phi(z) , dz$$
and $langle z,T^-1y rangle = langle (T^-1)^tz,y rangle = langle (T^t)^-1z,y rangle$, where the last equality uses the fact that the matrix representing $T$ is real.

Now combine the formulas. I will be a bit cavalier with the use of Fubini's theorem trusting that you can make it precise.

beginalign* widehatT^*u(phi) &= |det T|^-1 int_mathbb R^n int_mathbb R^n u(y) e^-i langle (T^t)^-1z,y rangle phi(z) , dz dy \ &= |det T|^-1 int_mathbb R^n int_mathbb R^n e^-i langle (T^t)^-1z,y rangle u(y), dy phi(z) , dz.
endalign*

As long as $u in L^1(mathbb R^n)$ (which doesn't seem to be implied by the hypotheses, but whatever) the inner integral equals
$ hat u ((T^t)^-1z) $ so that
$$widehatT^*u(phi) = |det T|^-1 int_mathbb R^n hat u ((T^t)^-1z) phi(z) , dz = |det T|^-1 int_mathbb R^n ((T^t)^-1)^* hat u(z) phi(z) , dz$$ where the last expression is
$|det T|^-1 ((T^t)^-1)^* hat u (phi).$ Finally disregard $phi$ to obtain
$$widehatT^*u = |det T|^-1 ((T^t)^-1)^* hat u.$$

share|cite|improve this answer

answered Jul 18 at 15:22

Umberto P.

34.6k12859

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome!! Since I am not familiar with Fourier transform, I find out that I misapplied the formula $ int_mathbbR^nhatf(x)g(x)dx= int_mathbbR^nf(x)hatg(x)dx $ which led to the dead end. Thanks a lot~
  – Yuchen
  Jul 18 at 15:46
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let us start with showing that $hat phi circ T^-1 = |det(T)| , widehatphi circ T^t$:
$$
(hat phi circ T^-1)(xi)
= hat phi (T^-1xi)
= int phi(x) , e^-ilangle x, , T^-1 xi rangle , dx
= int phi(x) , e^-ilangle (T^-1)^t x, , xi rangle , dx \
= y = (T^-1)^t x
= int phi(((T^-1)^t)^-1 x) , e^-ilangle y, , xi rangle , |det((T^-1)^t)^-1| , dy \
= |det(T)| int phi(T^t x) , e^-ilangle y, , xi rangle , , dy
= |det(T)| , widehatphi circ T^t(xi)
$$

Now, we can show what you want to show. I use a different notation than you; instead of $u(phi)$ I write $langle u, phirangle.$
$$
langle widehatT^* u, phi rangle
= langle T^* u, hat phi rangle
= langle u, |det T|^-1 hat phi circ T^-1 rangle
= |det T|^-1 langle u, hat phi circ T^-1 rangle \
= text by what was shown above \
= langle u, widehatphi circ T^t rangle
= langle hat u, phi circ T^t rangle
= |det(T^-1)| , langle ((T^t)^-1)^* hat u, phi rangle
$$

share|cite|improve this answer

answered Jul 18 at 15:26

md2perpe

5,93011022

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You were off to a good start. For any $phi in cal S(mathbb R^n)$ the definitions provide
$$widehatT^*u(phi) = int_mathbb R^n T^*u(x) hat phi(x) , dx = int_mathbb R^n u(Tx) hat phi(x), dx.$$

Now change variables: with $y = Tx$ you get $dy = |det T| dx$ so that
$$ int_mathbb R^n u(Tx) hat phi(x), dx = |det T|^-1 int_mathbb R^n u(y) hat phi (T^-1y) , dy.$$

Time to investigate the Fourier transform. Observe
$$ hat phi (T^-1 y) = int_mathbb R^n e^-i langle z,T^-1y rangle phi(z) , dz$$
and $langle z,T^-1y rangle = langle (T^-1)^tz,y rangle = langle (T^t)^-1z,y rangle$, where the last equality uses the fact that the matrix representing $T$ is real.

Now combine the formulas. I will be a bit cavalier with the use of Fubini's theorem trusting that you can make it precise.

beginalign* widehatT^*u(phi) &= |det T|^-1 int_mathbb R^n int_mathbb R^n u(y) e^-i langle (T^t)^-1z,y rangle phi(z) , dz dy \ &= |det T|^-1 int_mathbb R^n int_mathbb R^n e^-i langle (T^t)^-1z,y rangle u(y), dy phi(z) , dz.
endalign*

As long as $u in L^1(mathbb R^n)$ (which doesn't seem to be implied by the hypotheses, but whatever) the inner integral equals
$ hat u ((T^t)^-1z) $ so that
$$widehatT^*u(phi) = |det T|^-1 int_mathbb R^n hat u ((T^t)^-1z) phi(z) , dz = |det T|^-1 int_mathbb R^n ((T^t)^-1)^* hat u(z) phi(z) , dz$$ where the last expression is
$|det T|^-1 ((T^t)^-1)^* hat u (phi).$ Finally disregard $phi$ to obtain
$$widehatT^*u = |det T|^-1 ((T^t)^-1)^* hat u.$$

share|cite|improve this answer

answered Jul 18 at 15:22

Umberto P.

34.6k12859

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome!! Since I am not familiar with Fourier transform, I find out that I misapplied the formula $ int_mathbbR^nhatf(x)g(x)dx= int_mathbbR^nf(x)hatg(x)dx $ which led to the dead end. Thanks a lot~
  – Yuchen
  Jul 18 at 15:46
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

You were off to a good start. For any $phi in cal S(mathbb R^n)$ the definitions provide
$$widehatT^*u(phi) = int_mathbb R^n T^*u(x) hat phi(x) , dx = int_mathbb R^n u(Tx) hat phi(x), dx.$$

Now change variables: with $y = Tx$ you get $dy = |det T| dx$ so that
$$ int_mathbb R^n u(Tx) hat phi(x), dx = |det T|^-1 int_mathbb R^n u(y) hat phi (T^-1y) , dy.$$

Time to investigate the Fourier transform. Observe
$$ hat phi (T^-1 y) = int_mathbb R^n e^-i langle z,T^-1y rangle phi(z) , dz$$
and $langle z,T^-1y rangle = langle (T^-1)^tz,y rangle = langle (T^t)^-1z,y rangle$, where the last equality uses the fact that the matrix representing $T$ is real.

Now combine the formulas. I will be a bit cavalier with the use of Fubini's theorem trusting that you can make it precise.

beginalign* widehatT^*u(phi) &= |det T|^-1 int_mathbb R^n int_mathbb R^n u(y) e^-i langle (T^t)^-1z,y rangle phi(z) , dz dy \ &= |det T|^-1 int_mathbb R^n int_mathbb R^n e^-i langle (T^t)^-1z,y rangle u(y), dy phi(z) , dz.
endalign*

As long as $u in L^1(mathbb R^n)$ (which doesn't seem to be implied by the hypotheses, but whatever) the inner integral equals
$ hat u ((T^t)^-1z) $ so that
$$widehatT^*u(phi) = |det T|^-1 int_mathbb R^n hat u ((T^t)^-1z) phi(z) , dz = |det T|^-1 int_mathbb R^n ((T^t)^-1)^* hat u(z) phi(z) , dz$$ where the last expression is
$|det T|^-1 ((T^t)^-1)^* hat u (phi).$ Finally disregard $phi$ to obtain
$$widehatT^*u = |det T|^-1 ((T^t)^-1)^* hat u.$$

share|cite|improve this answer

answered Jul 18 at 15:22

Umberto P.

34.6k12859

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome!! Since I am not familiar with Fourier transform, I find out that I misapplied the formula $ int_mathbbR^nhatf(x)g(x)dx= int_mathbbR^nf(x)hatg(x)dx $ which led to the dead end. Thanks a lot~
  – Yuchen
  Jul 18 at 15:46
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You were off to a good start. For any $phi in cal S(mathbb R^n)$ the definitions provide
$$widehatT^*u(phi) = int_mathbb R^n T^*u(x) hat phi(x) , dx = int_mathbb R^n u(Tx) hat phi(x), dx.$$

Now change variables: with $y = Tx$ you get $dy = |det T| dx$ so that
$$ int_mathbb R^n u(Tx) hat phi(x), dx = |det T|^-1 int_mathbb R^n u(y) hat phi (T^-1y) , dy.$$

Time to investigate the Fourier transform. Observe
$$ hat phi (T^-1 y) = int_mathbb R^n e^-i langle z,T^-1y rangle phi(z) , dz$$
and $langle z,T^-1y rangle = langle (T^-1)^tz,y rangle = langle (T^t)^-1z,y rangle$, where the last equality uses the fact that the matrix representing $T$ is real.

Now combine the formulas. I will be a bit cavalier with the use of Fubini's theorem trusting that you can make it precise.

beginalign* widehatT^*u(phi) &= |det T|^-1 int_mathbb R^n int_mathbb R^n u(y) e^-i langle (T^t)^-1z,y rangle phi(z) , dz dy \ &= |det T|^-1 int_mathbb R^n int_mathbb R^n e^-i langle (T^t)^-1z,y rangle u(y), dy phi(z) , dz.
endalign*

As long as $u in L^1(mathbb R^n)$ (which doesn't seem to be implied by the hypotheses, but whatever) the inner integral equals
$ hat u ((T^t)^-1z) $ so that
$$widehatT^*u(phi) = |det T|^-1 int_mathbb R^n hat u ((T^t)^-1z) phi(z) , dz = |det T|^-1 int_mathbb R^n ((T^t)^-1)^* hat u(z) phi(z) , dz$$ where the last expression is
$|det T|^-1 ((T^t)^-1)^* hat u (phi).$ Finally disregard $phi$ to obtain
$$widehatT^*u = |det T|^-1 ((T^t)^-1)^* hat u.$$

share|cite|improve this answer

answered Jul 18 at 15:22

Umberto P.

34.6k12859

You were off to a good start. For any $phi in cal S(mathbb R^n)$ the definitions provide
$$widehatT^*u(phi) = int_mathbb R^n T^*u(x) hat phi(x) , dx = int_mathbb R^n u(Tx) hat phi(x), dx.$$

Now change variables: with $y = Tx$ you get $dy = |det T| dx$ so that
$$ int_mathbb R^n u(Tx) hat phi(x), dx = |det T|^-1 int_mathbb R^n u(y) hat phi (T^-1y) , dy.$$

Time to investigate the Fourier transform. Observe
$$ hat phi (T^-1 y) = int_mathbb R^n e^-i langle z,T^-1y rangle phi(z) , dz$$
and $langle z,T^-1y rangle = langle (T^-1)^tz,y rangle = langle (T^t)^-1z,y rangle$, where the last equality uses the fact that the matrix representing $T$ is real.

Now combine the formulas. I will be a bit cavalier with the use of Fubini's theorem trusting that you can make it precise.

beginalign* widehatT^*u(phi) &= |det T|^-1 int_mathbb R^n int_mathbb R^n u(y) e^-i langle (T^t)^-1z,y rangle phi(z) , dz dy \ &= |det T|^-1 int_mathbb R^n int_mathbb R^n e^-i langle (T^t)^-1z,y rangle u(y), dy phi(z) , dz.
endalign*

As long as $u in L^1(mathbb R^n)$ (which doesn't seem to be implied by the hypotheses, but whatever) the inner integral equals
$ hat u ((T^t)^-1z) $ so that
$$widehatT^*u(phi) = |det T|^-1 int_mathbb R^n hat u ((T^t)^-1z) phi(z) , dz = |det T|^-1 int_mathbb R^n ((T^t)^-1)^* hat u(z) phi(z) , dz$$ where the last expression is
$|det T|^-1 ((T^t)^-1)^* hat u (phi).$ Finally disregard $phi$ to obtain
$$widehatT^*u = |det T|^-1 ((T^t)^-1)^* hat u.$$

share|cite|improve this answer

answered Jul 18 at 15:22

Umberto P.

34.6k12859

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 15:22

Umberto P.

34.6k12859

answered Jul 18 at 15:22

Umberto P.

34.6k12859

answered Jul 18 at 15:22

Umberto P.

34.6k12859

34.6k12859

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome!! Since I am not familiar with Fourier transform, I find out that I misapplied the formula $ int_mathbbR^nhatf(x)g(x)dx= int_mathbbR^nf(x)hatg(x)dx $ which led to the dead end. Thanks a lot~
  – Yuchen
  Jul 18 at 15:46
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome!! Since I am not familiar with Fourier transform, I find out that I misapplied the formula $ int_mathbbR^nhatf(x)g(x)dx= int_mathbbR^nf(x)hatg(x)dx $ which led to the dead end. Thanks a lot~
  – Yuchen
  Jul 18 at 15:46
  
  

  
  
  
  

Awesome!! Since I am not familiar with Fourier transform, I find out that I misapplied the formula $ int_mathbbR^nhatf(x)g(x)dx= int_mathbbR^nf(x)hatg(x)dx $ which led to the dead end. Thanks a lot~
– Yuchen
Jul 18 at 15:46

Awesome!! Since I am not familiar with Fourier transform, I find out that I misapplied the formula $ int_mathbbR^nhatf(x)g(x)dx= int_mathbbR^nf(x)hatg(x)dx $ which led to the dead end. Thanks a lot~
– Yuchen
Jul 18 at 15:46

add a comment | 

up vote
0
down vote

Let us start with showing that $hat phi circ T^-1 = |det(T)| , widehatphi circ T^t$:
$$
(hat phi circ T^-1)(xi)
= hat phi (T^-1xi)
= int phi(x) , e^-ilangle x, , T^-1 xi rangle , dx
= int phi(x) , e^-ilangle (T^-1)^t x, , xi rangle , dx \
= y = (T^-1)^t x
= int phi(((T^-1)^t)^-1 x) , e^-ilangle y, , xi rangle , |det((T^-1)^t)^-1| , dy \
= |det(T)| int phi(T^t x) , e^-ilangle y, , xi rangle , , dy
= |det(T)| , widehatphi circ T^t(xi)
$$

Now, we can show what you want to show. I use a different notation than you; instead of $u(phi)$ I write $langle u, phirangle.$
$$
langle widehatT^* u, phi rangle
= langle T^* u, hat phi rangle
= langle u, |det T|^-1 hat phi circ T^-1 rangle
= |det T|^-1 langle u, hat phi circ T^-1 rangle \
= text by what was shown above \
= langle u, widehatphi circ T^t rangle
= langle hat u, phi circ T^t rangle
= |det(T^-1)| , langle ((T^t)^-1)^* hat u, phi rangle
$$

share|cite|improve this answer

answered Jul 18 at 15:26

md2perpe

5,93011022

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up vote
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Let us start with showing that $hat phi circ T^-1 = |det(T)| , widehatphi circ T^t$:
$$
(hat phi circ T^-1)(xi)
= hat phi (T^-1xi)
= int phi(x) , e^-ilangle x, , T^-1 xi rangle , dx
= int phi(x) , e^-ilangle (T^-1)^t x, , xi rangle , dx \
= y = (T^-1)^t x
= int phi(((T^-1)^t)^-1 x) , e^-ilangle y, , xi rangle , |det((T^-1)^t)^-1| , dy \
= |det(T)| int phi(T^t x) , e^-ilangle y, , xi rangle , , dy
= |det(T)| , widehatphi circ T^t(xi)
$$

Now, we can show what you want to show. I use a different notation than you; instead of $u(phi)$ I write $langle u, phirangle.$
$$
langle widehatT^* u, phi rangle
= langle T^* u, hat phi rangle
= langle u, |det T|^-1 hat phi circ T^-1 rangle
= |det T|^-1 langle u, hat phi circ T^-1 rangle \
= text by what was shown above \
= langle u, widehatphi circ T^t rangle
= langle hat u, phi circ T^t rangle
= |det(T^-1)| , langle ((T^t)^-1)^* hat u, phi rangle
$$

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answered Jul 18 at 15:26

md2perpe

5,93011022

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let us start with showing that $hat phi circ T^-1 = |det(T)| , widehatphi circ T^t$:
$$
(hat phi circ T^-1)(xi)
= hat phi (T^-1xi)
= int phi(x) , e^-ilangle x, , T^-1 xi rangle , dx
= int phi(x) , e^-ilangle (T^-1)^t x, , xi rangle , dx \
= y = (T^-1)^t x
= int phi(((T^-1)^t)^-1 x) , e^-ilangle y, , xi rangle , |det((T^-1)^t)^-1| , dy \
= |det(T)| int phi(T^t x) , e^-ilangle y, , xi rangle , , dy
= |det(T)| , widehatphi circ T^t(xi)
$$

Now, we can show what you want to show. I use a different notation than you; instead of $u(phi)$ I write $langle u, phirangle.$
$$
langle widehatT^* u, phi rangle
= langle T^* u, hat phi rangle
= langle u, |det T|^-1 hat phi circ T^-1 rangle
= |det T|^-1 langle u, hat phi circ T^-1 rangle \
= text by what was shown above \
= langle u, widehatphi circ T^t rangle
= langle hat u, phi circ T^t rangle
= |det(T^-1)| , langle ((T^t)^-1)^* hat u, phi rangle
$$

share|cite|improve this answer

answered Jul 18 at 15:26

md2perpe

5,93011022

Let us start with showing that $hat phi circ T^-1 = |det(T)| , widehatphi circ T^t$:
$$
(hat phi circ T^-1)(xi)
= hat phi (T^-1xi)
= int phi(x) , e^-ilangle x, , T^-1 xi rangle , dx
= int phi(x) , e^-ilangle (T^-1)^t x, , xi rangle , dx \
= y = (T^-1)^t x
= int phi(((T^-1)^t)^-1 x) , e^-ilangle y, , xi rangle , |det((T^-1)^t)^-1| , dy \
= |det(T)| int phi(T^t x) , e^-ilangle y, , xi rangle , , dy
= |det(T)| , widehatphi circ T^t(xi)
$$

Now, we can show what you want to show. I use a different notation than you; instead of $u(phi)$ I write $langle u, phirangle.$
$$
langle widehatT^* u, phi rangle
= langle T^* u, hat phi rangle
= langle u, |det T|^-1 hat phi circ T^-1 rangle
= |det T|^-1 langle u, hat phi circ T^-1 rangle \
= text by what was shown above \
= langle u, widehatphi circ T^t rangle
= langle hat u, phi circ T^t rangle
= |det(T^-1)| , langle ((T^t)^-1)^* hat u, phi rangle
$$

share|cite|improve this answer

answered Jul 18 at 15:26

md2perpe

5,93011022

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 15:26

md2perpe

5,93011022

answered Jul 18 at 15:26

md2perpe

5,93011022

answered Jul 18 at 15:26

md2perpe

5,93011022

5,93011022

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add a comment | 

 

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