Tψ, one has
FtTψ[f]≡Tψ[Ff]
=Tψ
[
1
√
2 π
∫+∞
−∞
e−ikqf(q)dq
]
=
1
√
2 π
∫+∞
−∞
ψ(k)
(∫+∞
−∞
e−ikqf(q)dq
)
dk
=
∫+∞
−∞
(
1
√
2 π
∫+∞
−∞
e−ikqψ(k)dk
)
f(q)dq
=TFψ[f]
showing that the Fourier transform is compatible with this identification.
As an example, the Fourier transform of the distribution √^12 πeikais the
δ-functionδ(q−a) since
FtT√^1
2 πe
ika[f] =
1
√
2 π
∫+∞
−∞
eika
(
1
√
2 π
∫+∞
−∞
e−ikqf(q)dq
)
dk
=
∫+∞
−∞
(
1
2 π
∫+∞
−∞
e−ik(q−a)dk
)
f(q)dq
=
∫+∞
−∞
δ(q−a)f(q)dq
=Tδ(q−a)[f]
For another example of a linear transformation acting onS(R), consider the
translation action on functionsf→Aaf, where
(Aaf)(q) =f(q−a)
The transpose action on distributions is
AtaTψ(q)=Tψ(q+a)
since
AtaTψ(q)[f] =Tψ(q)[f(q−a)] =
∫+∞
−∞
ψ(q)f(q−a)dq=
∫+∞
−∞
ψ(q′+a)f(q′)dq′
The derivative is an infinitesimal version of this, and one sees (using inte-
gration by parts), that
(
d
dq
)t
Tψ(q)[f] =Tψ(q)
[
d
dq
f
]
=
∫+∞
−∞
ψ(q)
d
dq
f(q)dq
=
∫+∞
−∞
(
−
d
dq
ψ(q)
)
f(q)dq
=T−dqdψ(q)[f]