Is it possible to make sense, in distributional sense, of the Fourier transform of the exponential function (defined over the whole real line)?
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1$\begingroup$ Since $e^{iax}$ has Fourier transform $\delta_a$, we formally might think of taking $a = -i$, suggesting that the Fourier transform of $e^x$ "should be" $\delta_{-i}$, i.e. a functional taking $f$ to $f(-i)$. To me, this suggests that the right domain of test functions for such a functional should be holomorphic functions. It looks like the theory of hyperfunctions (en.wikipedia.org/wiki/Hyperfunction) treats this; I don't know anything about them, but it may be worth a look. $\endgroup$– Nate EldredgeOct 13, 2010 at 3:32
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1$\begingroup$ Hyperfunctions are an interesting hint for this question, but as far as I know the Fourier transform is defined for hyperfunctions of sub-exponential growth only (see Kaneko, Arscott: "Introduction int o hyperfunctions"). (A functional like $\delta_i$ is not a hyperfunction.) $\endgroup$– Tim van BeekOct 13, 2010 at 11:49
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2$\begingroup$ There's no need to introduce hyperfunctions -- Nate Eldredge's answer is already correct. A smooth test function of compact support has a Fourier transform which is naturally an entire function (defined by the same formula as the usual Fourier transform). Integrating against, say, $e^x$ will just give you the value at $-i$ as you can see from the formula (depending on your normalizations), just as you would expect. It already makes sense -- in my opinion, you don't honestly need to know the "correct" space of test functions until you run into a specific problem. $\endgroup$– Phil IsettAug 7, 2011 at 3:50
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6 Answers
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It is possible to make sense of this if one generalizes the notion of distribution by choosing a smaller space of test functions. The space of test functions should be chosen so that it is closed under Fourier transform and its elements decrease so fast that multiplying them by an exponential function is still integrable. Some possibilities for the space of test functions with these properties are:
*Holomorphic functions on the complex plane that decrease faster than any exponential on horizontal strips (if I have remembered the Paley-Wiener theorem correctly...)
*A more extreme space of test functions is polynomials times Gaussians.
(Using smooth compactly supported test functions as suggested in another answer does not work as this space is not closed under Fourier transforms)
answered Oct 13, 2010 at 13:50
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4$\begingroup$ The spaces you describe are, I think, the Gelfand-Shilov spaces $S^a_a$. For a Fourier characterization of them see jstor.org/stable/2161498 . I don't know much about them personally (having only skimmed a bit of literature); they are not particularly useful to me because they are either analytic or quasi-analytic (I forgot which), with the notable property that the test function space does not contain any function of compact support (in fact, I think they may not even have functions which vanish on open sets, as long as some condition on the index $a$ is obeyed.) $\endgroup$ Oct 13, 2010 at 14:45
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3$\begingroup$ To elaborate on why smooth compactly supported test functions fail: Using asymptotic methods, the Fourier transform of the (smooth compactly supported) function $\phi_a(x)=e^{-1/(a^2-x^2)}$ (and $\phi(x)=0$ for $|x|\ge a$) has exponential decay depending on $a$, so that for any exponential function $e^{\lambda x}$, we can find a $\phi_a$ so that $\int e^{\lambda x}\hat\phi_a(x)$ diverges. Thus the distributional Fourier transform of $e^{\lambda x}$ does not exist (because we can't integrate it against the Fourier transform of all test functions). $\endgroup$– B ROct 13, 2010 at 17:20
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$\begingroup$ Is there a reason why the Schwartz space is more widely acknowledged and popular? Do we lose something when moving to smaller sets of test functions (that are closed under Fourier transforms)? Having a Fourier transform of a real exponential seems like a significant achievement to me, so I do not understand why most sources do not extend the class of distributions beyond the tempered ones, unless there is a good reason not to do so. $\endgroup$– ZeeklessMar 24 at 19:34
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As commented above, $e^{tx}$ with t real is not a tempered distribution, so the simplest standard extension of Fourier transform to tempered distributions does not apply.
However, also as commented above, shrinking the space of Schwartz functions to something smaller, the extreme case being compactly-supported smooth, makes real exponentials continuous functionals, so Fourier transform can be defined. True, since Fourier transform does not map test functions to themselves (but to Paley-Wiener spaces), such extensions of Fourier transforms will map real exponentials to elements of the dual of such very-slightly-exotic spaces, rather than to spaces of distributions.
I think this is a natural outcome.
answered Jun 3, 2011 at 12:59
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The Fourier transforms of distributions (not necessarily of Schwartz class) form a space called analytic functionals. A detailed discussion of these can be found in Gelfand and Shilov.
answered Jun 3, 2011 at 13:36
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L. Ehrenpreis, in his series of papers on the division problem in the 1950's and in his book Fourier analysis in the complex domain, introduced and developed the Fourier transform of distributions. A Fourier transform of a distribution $u\in C^{-\infty}(\Omega)$, $\Omega$ an open, convex subset of euclidean $n$-space, is a Radon measure $\mu$ such that $$u(\varphi)=\int \hat\varphi(z)\,d\mu(z),\quad \varphi\in C_c^\infty(\Omega),$$ where $\hat\varphi$ is the Fourier-Laplace transform of the test function $\varphi$. The measure $\mu$ satisfies uniform bounds which derive from the Paley-Wiener estimates. (The Paley-Wiener theorem has been alluded to in some answers.) More precicely, Ehrenpreis developed his theory of analytically uniform (AU) spaces for the purpose of Fourier representation of general distributions. "The" Fourier transform $\mu$ of $u$ is not unique of course. Still, it is very useful in proving, e.g., the Ehrenpreis-Malgrange theorem and the Ehrenpreis Fundamental Principle about exponential representation of solutions to homogeneous linear partial differential system with constant coefficients. The Dirac measure at a complex number $\lambda$ is, of course, a Fourier transform of $u(t)=e^{i\lambda t}$.
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The Fourier transform of $f_{\lambda}\left(\omega\right)=2\sqrt{\pi\lambda}e^{\left(a+i\omega\right)^{2}\lambda}$ is $e^{as-s^{2}/\left(4\lambda\right)}$. This approximates the exponential function for $\left|s\right|\ll a\lambda$. In physics (where everything is finite) one is interested in the Fourier transform of the causal response function $e^{as}\theta(s)=e^{as}-e^{as}\theta(-s)$, approximated by $f_{\lambda}\left(\omega\right)-1/\left(a+i\omega\right)$ for $\left|s\right|\ll a\lambda$.
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Of course, yes. Given a real number $\lambda$ define the distribution $$ E_{\lambda}(u) = \int e^{i \lambda x} u(x) dx $$ for all test functions $u$ (that is $u$ is smooth and compactly supported). Then the Fourier transform of $E_{\lambda}(u)$ is defined by $$ \hat{E_{\lambda}}(u) = E_{\lambda}(\hat{u}) $$ which you can check to be equal to $\hat{u }(\lambda)$.
The interpretation is that you extend the equality $$ \int f(x) \hat{g}(x) dx = \int \hat{f}(x) g(x) dx $$ valid for all test functions $f,g$ to distributions.
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4$\begingroup$ The Fourier transform of $E_\lambda$, which is the complex exponential tempered distribution, is $\delta_\lambda$ (depending on normalization). Further, if the OP meant the real exponential function, as a distribution with exponential growth, the answer is almost surely no. The Fourier transform is initially defined on Schwartz functions. We can extend it to negatively indexed ($L^2$) Sobolev spaces, but I don't see a way to continue it to distributions in general. $\endgroup$– B ROct 13, 2010 at 4:37