Fourier transform of periodic distributions

Question

Following M. Ruzhansky and V. Turunen's book Pseudo-Differential Operators and Symmetries, in Chapter 3, Definition 3.1.25 (page 304), the space of periodic distributions is defined as follows (paraphrasing):

Definition 3.1.25 (Periodic Distributions) The space of periodic distributions is defined as dual space $\mathcal{D}'(\mathbb{T}^n) = \mathcal{L}(C^{\infty}(\mathbb{T}^n),\mathbb{C})$ (i.e. of continuous linear operators $u:C^{\infty}(\mathbb{T}^n) \to \mathbb{C}$), where we write $u(\varphi) = \langle u , \varphi\rangle$ for any $\varphi\in C^{\infty}(\mathbb{T}^n)$. Furthermore, for any $\psi\in C^\infty(\mathbb{T}^n)$ or $L^p(\mathbb{T}^n)$, define the associated distribution to $\psi$ in the canonical sense, i.e. $u_{\psi} \in \mathcal{D}'(\mathbb{T}^n)$ where for any $\varphi \in C^\infty(\mathbb{T^n})$ \begin{align}u_{\psi}(\varphi) = \langle u_{\psi}, \varphi\rangle=\int_{\mathbb{T}^n}\psi(x) \varphi(x) dx.\end{align}

Given the above definition, as with standard distributions, we can define a Fourier transform of periodic distributions with respect to Fourier transforms of functions.

Definition 3.1.8 (Periodic Fourier Transform) Let $\mathcal{F}_{\mathbb{T}^n} : C^{\infty}(\mathbb{T^n}) \to \mathcal{S}(\mathbb{Z}^n) $ be defined as $$ \begin{align} (\mathcal{F}_{\mathbb{T}^n}f)(\xi):=\int_{\mathbb{T}^n} f(x) e^{-2\pi i x \cdot \xi} dx \end{align}$$ for $f \in C^{\infty}({\mathbb{T}^n})$, and its inverse be defined as $$\begin{align} (\mathcal{F}_{\mathbb{T}^n}^{-1}h)(x) = \sum_{\xi \in \mathbb{Z}^n} h(\xi) e^{2\pi i x \cdot \xi}\end{align}$$ for $h\in\mathcal{S}({\mathbb{Z}^n})$.

Definition 3.1.27 (Fourier Transform of Periodic Distributions) For any $u\in\mathcal{D}'(\mathbb{T}^n)$, define the Fourier transform $\mathcal{F}_{\mathbb{T}^n}:\mathcal{D}'(\mathbb{T}^n) \to \mathcal{S}'(\mathbb{Z}^n)$ as $$\begin{align}\langle \mathcal{F}_{\mathbb{T}^n} u, \varphi \rangle = \langle u, \mathcal{F}_{\mathbb{T}^n}^{-1} \varphi(-\cdot) \rangle\end{align} \tag{1}$$ for any $\varphi \in C^{\infty}(\mathbb{T}^n)$

QUESTION

My question is, is the definition of the Fourier transform on periodic distributions, as given in Definition 3.1.27, correct? I feel that the definition should not involve the inverse Fourier transform, but instead be defined as $$\begin{align} \langle \mathcal{F}_{\mathbb{T}^n} u, \varphi \rangle = \langle u, \mathcal{F}_{\mathbb{T}^n} \varphi \rangle. \tag{2} \end{align}$$

Reason

The reason I feel that this should be the case is because, if we consider the periodic Dirac distribution at zero, defined as $\delta_0 \in \mathcal{D}'(\mathbb{T}^n)$ where $$ \begin{align}\langle \delta_0, \varphi\rangle = \varphi(0)\end{align} $$ for any $\varphi \in C^{\infty}(\mathbb{T}^n)$, then its Fourier transform using $(1)$ gives $$ \begin{align} \underbrace{\langle \mathcal{F}_{\mathbb{T}^n} \delta_0, \varphi \rangle}_{\in \mathbb{C}} &= \langle \delta_0, \mathcal{F}_{\mathbb{T}^n}^{-1} \varphi (-\cdot) \rangle \\ &= \mathcal{F}_{\mathbb{T}^n}^{-1} \varphi (0) \\ &= \sum_{\xi \in \mathbb{Z}^n} \varphi(\xi), \end{align} $$ where the right hand side is infinite since $\varphi$ is periodic. So this doesn't make sense. If we instead use $(2)$, then we obtain the following: $$ \begin{align} \underbrace{\langle \mathcal{F}_{\mathbb{T}^n} \delta_0, \varphi \rangle}_{\in \mathbb{C}} &= \langle \delta_0, \mathcal{F}_{\mathbb{T}^n} \varphi \rangle \\ &= (\mathcal{F}_{\mathbb{T}^n} \varphi) (0)\\ &= \int_{\mathbb{T}^n} \varphi(x) dx \\ &= \langle 1, \varphi \rangle \end{align} $$ i.e. $\mathcal{F}_{\mathbb{T}^n} \delta_0 = 1$ in the sense of periodic distributions, as we would expect.

Answer 1

Actually, the definition you gave in the post differs from the one in the book. The test function $\varphi$ should lie in $\mathcal S(\mathbb Z^n)$, not in $C^\infty(\mathbb T^n)$, since the $\mathcal F_{\mathbb T^n}$ maps the second of these spaces into the first, so for $\varphi\in C^\infty(\mathbb T^n)$ the expression $\mathcal F_{\mathbb T^n}^{-1}\varphi$ is undefined, as you correctly noticed. So, the correct formula would be $$ \langle\mathcal F_{\mathbb T^n}u,\varphi \rangle=\langle u,\iota\,\circ\,\mathcal F_{\mathbb T^n}^{-1}\varphi \rangle, $$ where $\iota(f)(x)=f(-x)$ and $\varphi \in \mathcal S(\mathbb Z^n)$.

Answer 2

This is a comment rather than an answer but it will be too long. There is a confusion in your statement which I find rather irritating and which has not, as far as I can see, been addressed here. It is manifested in the formula where you write $\phi(0)$ despite the fact that $\phi$ is a smooth function on the torus. Here is my take:

1. The periodic distributions can be visualised in two (mathematically equivalent) ways: as distributions on the line which satisfy the usual periodicity condition or as distributions on the circle (= $1$-dimensional torus). (For simplicity, I will confine myself to the one-dimensional case). Under this identification, the $\delta$-distribution at the point $1$ on the circle (embedded into the plane) corresponds to the periodic $\delta$-distribution on the line.

2. Every distribution on the torus has a Fourier series, with coefficients in the space of slowly increasing sequences indexed by the whole numbers (usually written $s´$ rather than $\cal S´$ as you write. (This corresponds to the group duality between the line and the whole numbers).

3. Every periodic distribution on the line is a tempered distribution and so has a Fourier transform (which is also a periodic (tempered) distribution). In terms of harmonic analysis, this corresponds to the self-duality of the line.

4. In a certain sense these two F.T.´s coincide--thus the periodic delta distribution on the line has F.T. $\sum_n a_n e^{2 \pi i n x}$ whereas its version on the torus has Fourier series $\sum a_n z^n$ (both sums over the whole numbers--we use the complex variable on the torus to avoid confusion but see below), up to the usual constant. For example the case of the delta function on the torus, centred at the complex number $1$, is the case where all of the coefficients are $1$, exactly as for its disguise as the periodic delta function. Note that the fact that the latter is an infinite sum is of no consequence since if we regard it is a functional, then there are no convergence problems--it is applied to functions of rapid decrease.

5. All of this can be found in Schwartz´ original monograph. The fact that the Fourier series on the torus looks like a Laurent series is of some significance and was examined in detail by Köthe in his seminal work on distributions as boundary values of holomorphic functions.

Answer 3

In addition to other good comments and answers:

There is indeed a technical issue, which perhaps has scant intuitive content, but can matter regarding precision. This is touched upon in other comments/answers, but which I think deserves clarification. Namely, while "periodic distributions" on $\mathbb R$ are those which descend to the circle, and, oppositely, distributions on the circle compose with the projection of the line to the circle, giving "periodic distributions", there is a hitch with test functions or Schwartz functions.

The obvious, but also obviously harmless, point is that a smooth function on the circle, pulled back to the line, is not Schwartz. So it is possible to give flawed definitions of "periodic distributions", if one tries hard enough.

But/and the truly relevant point is that (already) the averaging map from test functions on $\mathbb R$ to smooth functions on the circle, by $f\to \sum_n f(x+n)$ is a surjection. This is what actually does identify periodic distributions with distributions on the circle. (Yes, we can notate this with suitable symbols.)

A sort of marginally-relevant artifact is that the Fourier transform of a periodic function (under mild hypotheses) is a periodic distribution supported on integers (or $2\pi$ times, etc.) That is, "Fourier series" of periodic functions, at least as tempered distributions, are also Fourier transforms... So, yes, everything is compatible. Yes, it's easy to say literally incorrect things. Mercifully, the stuff works really well. :)