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From: ALittleDog on 26 Mar 2010 06:37
By the way, is f(t) = c (constant) a square summable function.
Its Fourier transform does exit in Mathematica 7 FourierTransform[1,
t, \[Omega]] gives Sqrt[2 \[Pi]] DiracDelta[\[Omega]]

From: ALittleDog on 27 Mar 2010 06:09
Following the previous question, may it be said that a periodic has
its Fourier transform and a non-periodic function has its Fourier
transformed iff it is square integrable.

From: dh on 27 Mar 2010 06:08
Hi,
=>Is Exp[nt] also function basis of L2?

As Exp[n t] does not belong to L2 it can not be a basis.

=> f(t) = c (constant) a square summable function.
Its Fourier transform does exit in Mathematica 7 FourierTransform[1,
t, \[Omega]] gives Sqrt[2 \[Pi]] DiracDelta[\[Omega]]

A constant does not belong to L2(-Infinity,Infinity). However one can
generalize the notion of function to define e.g. Fourier transforms of a
constant. The generalized function only make sense inside an integral.
The Fourier integral of const is zero with the expcetion of \omega==0.
This is different from Exp[nt] that is unbounded.

Daniel



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From: ALittleDog on 27 Mar 2010 06:09
I got the following answers from Daniels:
" A constant does not belong to L2(-Infinity,Infinity). However one
can generalize the notion of function to define e.g. Fourier
transforms of a constant. The generalized function only make sense
inside an integral. The Fourier integral of const is zero with the
expcetion of \omega==0. This is different from Exp[nt] that is
unbounded. "

I guess Daniels' answer also applies to Cos[t], which is also not
integral summable and thus does not belong to L2.

However, the Cos[t] function is square-integrable of [ - pi,pi]. And a
constant can also be interpreted in this way to be square integrable
of [-pi, pi].
In contrast to that, Exp[-t] is also square integrable of [-pi,pi].
Then, what makes the difference between a constant or Cos[t] and
Exp[t], being not square integrable properties L2(-
Infinity,Infinity) ? Is it due to the periodicity of Cos[t] and a
constant?

From: Kevin J. McCann on 27 Mar 2010 06:08
Yes, Exp[I n t] is a basis for Hilbert space L2, but this means for
square integrable functions, and exp(-t) is not one of those.

Kevin

ALittleDog wrote:
> In fact, I want that t runs from minus infinity to infinity.
> But why Mathematica 5.0 gives \sqrt(2 Pi) DiracDelta[ I + \omega ] as
> the answer, when I run the code FourierTransform[Exp[-t], t, \omega],
> if the integral doesn't converge? Is there a different interpretation?
> Moreover, Exp[I n t] (n from minus infinity to positive infinity) are
> function basis of infinite number in Hilbert space L2. Is Exp[nt] also
> function basis of L2?
>
>

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