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Is there one-sided Fourier transform (unilateral Fourier transform)?

For example we have two-sided and one-sided Laplace transform.

living being
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  • I don't think so. By the convolution theorem, the one-sided FT would be the convolution of the transform of the given function with the transform of the Heaviside step (a Dirac delta and an inverse). –  Nov 18 '14 at 09:40

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If by "exists", you mean "can be defined", then one can of course define an operator $T$ on $L^1(\mathbb{R}^+)$ by

$$ T(f)(\xi) = \int_0^{\infty} e^{-2 \pi i x \xi} f(x) \, dx$$

I have seen integrals of this type arise in some applications, though I do not exactly remember the context. It should be usable, for example, to solve certain differential equations, since it behaves very similarly to the Laplace transform.

However, regardless of how it is used, the operator $T$ as defined above will not behave like the Fourier transform. The Fourier transform is fundamentally tied to convolutions, which only can be defined if the underlying space is a group; for the Euclidean Fourier transform this group is $(\mathbb{R}^n,+)$. The underlying space for $T$ is $(\mathbb{R}^+,+)$, which is merely a semigroup, not a group.

Christopher A. Wong
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