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y
x = 0, and the supposed Fourier transform is not in L 2 , because the integral of
its square is not finite. Nonetheless, with some care one can work with the delta
function by constantly remembering that the ordinary rules don't actually apply
to it, and it's really a proxy for a limiting process.
2
y = g ( x , 0.5)
1.5
function, or the comb
function defined in the next section, is inside a mapping like
The only way in which we'll ever want to use the
δ
1
y
=
g ( x , 1)
y
=
g ( x , 2)
0.5
f
R δ
( x ) f ( x ) dx ,
(18.65)
0
x
2
0
2
that is, to define a real-valued function on L 2 , which is a covector for L 2 .Ifwe
replace
0, the resultant
sequence of covectors actually does converge, although this requires proof. Thus,
the
δ
( x ) in Equation 18.65 with g ( x , a ) , and take a limit as a
Figure 18.46: x
g ( x , a ) for
several values of a.
δ
function, at least inside an integral, makes some sense.
y
1
18.15.2 The Comb Function and Its Transform
In much the same way we just analyzed a sequence of ever-narrower-and-taller
box functions, we can consider a sequence of L 2 ( R ) functions that approaches a
comb, a function with an “infinitely narrow box” at every integer. Figure 18.48
shows how we can do this: We place boxes of width a and height 1
0
x
4
2
0
2
4
Figure 18.47: The Fourier trans-
forms of the examples in Fig-
ure 18.46, matched by color.
/
a at each
integer point, but then multiply their heights by a “tapering” function of width
proportional to 1
a so that the total area under all the boxes is finite, hence the
functions are all in L 2 ( R ) .
Figure 18.49 shows the transforms of these functions. Just as with the delta
function, the transforms seem to approach a limit, but in this case the limit is
again the comb function (i.e., the transform grows larger and larger at integer
points, while heading toward zero at all noninteger points).
We'll use the symbol
/
1
0
.
If we create a comb with spacing c instead of 1, its transform is a comb with
spacing 1
ψ
for the comb; informally, we say that
F
(
ψ
)=
ψ
−5
0
5
/
c , just as we saw with the box and the sinc.
1
0
18.16 The Inverse Fourier Transform
5
0
5
We've already said that if we take a function f in L 2 ( H ) (or a periodic function of
period one) and compute its Fourier transform c k =
F
( f )( k ) , then we can recover
1
f by writing
0
5
0
5
c k e k ( t ) .
(18.66)
Figure
18.48:
Functions
that
k
approach a comb.
For a nice function f , this sum equals f except at points of discontinuity of f , and
possibly the endpoints, if f ( 2 )
1
2 ) . Thus, we've defined an inverse trans-
form that takes a sequence of coefficients and produces an L 2
= f (
function on the
interval (or a periodic function of period one).
There's a similar “inverse transform” defined for L 2 ( R ) :
F 1 ( g )( x )= C
−∞
g (
ω
) e ω ( x ) d
ω
,
(18.67)
 
 
 
 
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