Clarke. .Fourier.Theory.[sharethefiles.com]

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Let f*(x) a" ((i� f^(� ))� or (f*)^(� ) = i� f^(� ). Then f* " L2(R), and
" " " "
�� = ��
dx�� dx��
��
+" x2lf(x)l2 �� +" � 2lf^(� )l2d� �� +" x2lf(x)l2 �� +" l(i� )f^(� )l2d� ��
��
" " " "
" "
��
= �� dx��
��
+" x2lf(x)l2 �� +" l(f*)^(� )l2d� ��
��
" "
" "
��
= 2� dx�� dx��
��
+" x2lf(x)l2 �� +" lf*(x)l2 �� (by Parseval s identity).
��
" "
Since
" "
2 2
��
1
* *
�� x (x) f(x) + f*(x) f(x) �� = �� x Re (x) f(x) ��
(f )dx�� (f )dx��
+" 2 +"
��
" "
" "
2 2
��
= �� Re x f(x) f*(x) dx �� d" �� x f(x) f*(x) dx ��
+" +"
��
" "
" "
d" �� dx�� dx��
��
+" x2lf(x)l2 �� +" lf*(x)l2 ��
��
" "
(by the Cauchy-Schwartz inequality).
We will show that
" "
x (x) f(x) + f*(x) f(x) dx
(f )dx =
+" * +" lf(x)l2
" "
from which the result follows, for then
" " " "
�� = 2� ��
dx�� dx�� dx��
��
+" x2lf(x)l2 �� +" � 2lf^(� )l2d� �� +" x2lf(x)l2 �� +" lf*(x)l2 ��
��
" " " "
"
2
��
1
*
e" 2� �� x (x) f(x) + f*(x) f(x) ��
(f )dx��
+" 2
��
"
18.�Heisenberg�s�Inequality.
61
" "
�� 2 � �� "
�� 1 ��
= � �� dx�� dx��
��
2 +" lf(x)l2 �� = 2 +" lf(x)l2 �� +" lf^(� )l2d� ��
2�
��
" " "
" "
1 �� .
= dx��
��
4 +" lf(x)l2 �� +" lf^(� )l2d� ��
��
" "
To complete the proof, we assume that f is continuous and piecewise smooth, This
assumption can be removed since functions in L1(R) are the uniform limit of such
functions.
Then from the property of Fourier transforms,
f*(x) = f'(x)
wherever the derivative exists. Then for any interval [a, b] ,
b
d
b lf(b)l2 a lf(a)l2 =
+" dx (x lf(x)l2)dx
a
b
=
(x f '(x) f(x) + x f(x) f '(x) + lf(x)l2 )dx
+"
a
b b
= x (x) f(x) + f*(x) f(x) dx
(f )dx +
+" * +" lf(x)l2
a a
The assumption f " L2(R) implies that b lf(b)l2 �! 0 as b �! " and a lf(a)l2 �! 0 as a �! "
since otherwise lf(x)l > c lxl 1/2 as lxl �! " , which is not integrable. Taking the limit as
b �! " and a �! " ,
" "
0 = x (x) f(x) + f*(x) f(x) dx
(f )dx +
+" * +" lf(x)l2
" "
as required.
As for the case of equality in Heisenberg s inequality, this holds if and only if f(x) f*(x) is
real and f*(x) = K x f(x) for some complex constant K. That is,
f(x) f*(x) = f(x) K x f(x) = x lf(x)l2 K is real.
Therefore K is real.
The differential equation
f'(x) = f (x) = K x f(x)
has solutions of the form
Kx2

2
f(x) = c e , c any real constant,
18.�Heisenberg�s�Inequality.
62
Kx2

2
and f(x) = c e " L2(R) if and only if K > 0. Therefore equality holds in Heisenberg s
2
inequality only if f(x) = c e kx for constants c " R and k > 0.
Kx2

2
Conversely, let f(x) = e for constant K > 0. Then
" " " "
�� = 2� ��
dx�� dx�� dx��
��
+" x2lf(x)l2 �� +" � 2lf^(� )l2d� �� +" x2lf(x)l2 �� +" lf*(x)l2 ��
��
" " " "
" "
��
= 2� dx�� dx��
��
+" x2lf(x)l2 �� +" lK x f(x)l2 ��
��
" "
"
2
= 2� K2 �� dx��
��
+" x2lf(x)l2 ��
��
"
"
Kx2 2
= 2� K2 ��
��
+" x2e dx��
��
"
"
2
��
x
Kx2
( 2Kxe
)
= 2� K2 �� dx��
+" �� 2K��
��
��
"
"
2
��
x
Kx2
( 2Kxe
)
= 2� K2 �� dx��
+" �� 2K��
��
��
"
"
2
��
x d Kx2
( )��
= 2� K2 �� dx��
+" �� 2K�� dx e
��
��
"
"
2 2
��
1
= 2� K2 �� dx��
+" �� 2K�� e Kx �� (integration by parts)
��
��
"
"
� �� Kx2 2
��
= 2 ��
+" e dx��
��
"
"
� �� t2 2 2
��
= ��
2K +" e dt�� = 2K .
��
"
Whereas,
" " " "
1 �� = 1 ��
dx�� dx�� 2� dx��
��
4 +" lf(x)l2 �� +" lf^(� )l2d� �� 4 +" lf(x)l2 �� +" lf(x)l2 ��
�� " ��
" " "
"
2
�
= �� dx��
��
2 +" lf(x)l2 ��
��
"
18.�Heisenberg�s�Inequality.
63
"
� �� Kx2 2
��
= 2 ��
+" e dx��
��
"
and equality holds.
The case of a `" 0, � `" 0, follows by observing that F(x) = e i� x f(x + a) satisfies the same
hypotheses as f(x) and " f = " F and " f^= " F^ for any a `" 0, � `" 0.
a 0 � 0
As a consequence of the inequality f f^ 1
(" )(" � )e" , we see that it is impossible for
a 4
both " f and " f^ to be simultaneously small. That is, if one of " f or " f^ is very small
a � a �
then the other must be large. [ Pobierz całość w formacie PDF ]

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