It is usually stated that the Fourier Transform ($FT$) formulae are valid for functions which are absolutely integrable on the domain. Then we have formulae relating the $FT$ of the derivative with that of the function.

  1. If $\mathbf{x} \in \mathbb{R}^3$, then does it make sense to speak of the $FT$ of $\lVert{\mathbf{x}} \rVert$, seeing that $\lVert{\mathbf{x}} \rVert$ is not bounded ?

  2. Does the differentiation formula make sense for this case, that is $FT(f') = ikFT(f)$ ?

  • $\begingroup$ Does the term "tempered distribution" seem not entirely unheard of to you? $\endgroup$ – Daniel Fischer Oct 24 '16 at 12:34
  • $\begingroup$ @DanielFischer: I see it often come up but I dont understand it. Cuold you explain how that answers my question ? $\endgroup$ – me10240 Oct 24 '16 at 12:53

The term "Fourier Transform" denotes several (closely related) operators on different spaces. We start from the Fourier integral

$$\mathscr{F}_1[f](\omega) := \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} f(x) e^{- i \langle x\mid\omega\rangle}\,dx\tag{1}$$

defined for $f \in L^1(\mathbb{R}^n)$. [Note: There are various different normalisations of the Fourier transform in use. These different normalisations do not change anything essential, but introduce small differences in various formulae.] This defines $\mathscr{F}_1$ as a continuous linear operator $L^1(\mathbb{R}^n) \to C_0(\mathbb{R}^n)$ (where $C_0(X)$ is the space of continuous functions on $X$ "vanishing at infinity", endowed with the supremum norm).

This is however not the most useful setting for the Fourier transform. The image $\mathscr{F}_1(L^1(\mathbb{R}^n))$ is difficult to characterise - if a characterisation of the image is known, I'm not aware of it - and the inverse of $\mathscr{F}_1$ is not continuous.

One can use $\mathscr{F}_1$ to define the Fourier transform on $L^2(\mathbb{R}^n)$. One first restricts $(1)$ to the space $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$. Investigating that map, one sees that for $f \in L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ one has $\mathscr{F}_1[f] \in C_0(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, and - this is the incredibly useful thing -

$$\lVert \mathscr{F}_1[f]\rVert_{L^2(\mathbb{R}^n)} = \lVert f\rVert_{L^2(\mathbb{R}^n)}.\tag{2}$$

Thus $\mathscr{F}_1\lvert_{L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)} \colon L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n) \hookrightarrow L^2(\mathbb{R}^n)$ is an isometric injection. Since $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ is a dense subspace of $L^2(\mathbb{R}^n)$, there is a unique continuous linear extension $\mathscr{F}_2 \colon L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ of $\mathscr{F}_1\lvert_{L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)}$. By general principles this extension is still isometric, hence injective, and it turns out that it is also surjective. Thus $\mathscr{F}_2$ is an isometric automorphism of $L^2(\mathbb{R}^n)$. It is not surprising that this is a very useful property, and thus $L^2(\mathbb{R}^n)$ is one of the "natural homes" of the Fourier transform. Formulae like $\mathscr{F}[f\ast g] = (2\pi)^{n/2}\mathscr{F}[f]\cdot \mathscr{F}[g]$ or

$$\int_{\mathbb{R}^n} \mathscr{F}[f](\omega)g(\omega)\,d\omega = \int_{\mathbb{R}^n} f(x)\mathscr{F}[g](x)\,dx\tag{3}$$

(which are straightforward to prove for $\mathscr{F}_1$) continue to hold, as do the formulae involving differentiation (provided the derivatives exist in a suitable sense and satisfy the appropriate integrability conditions).

Clearly neither $\mathscr{F}_1$ nor $\mathscr{F}_2$ is applicable to $f(x) = \lVert x\rVert$. To deal with that, we first restrict the Fourier transform to a smaller space. Since the Fourier transform converts partial derivatives into multiplications with coordinate functions and vice versa, we can look at the space of smooth (infinitely differentiable) functions that remain integrable when multiplied with an arbitrary polynomial, and such that this also holds for all their derivatives. This gives us the Schwartz space $\mathscr{S}(\mathbb{R}^n)$ of "rapidly decreasing functions". We have $\mathscr{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n) \cap C_0(\mathbb{R}^n)$, and further $\mathscr{F}_1(\mathscr{S}(\mathbb{R}^n)) = \mathscr{S}(\mathbb{R}^n)$. Note that it is not particularly useful to endow $\mathscr{S}(\mathbb{R}^n)$ with the subspace topology induced by any $L^p(\mathbb{R}^n)$, since it is not complete in any of those norms (it is dense in $L^p$ for $p < \infty$, and its closure in $L^\infty$ is $C_0$). The "natural" topology on $\mathscr{S}(\mathbb{R}^n)$ is not normable, but it makes $\mathscr{S}(\mathbb{R}^n)$ a Fréchet space (more specifically, a Fréchet-Schwartz space), and the Fourier transform $\mathscr{F}_s := \mathscr{F}_1\lvert_{\mathscr{S}(\mathbb{R}^n)} \colon \mathscr{S}(\mathbb{R}^n) \to \mathscr{S}(\mathbb{R}^n)$ is continuous (and bijective), hence a topological automorphism.

Now we can make a big step and define the Fourier transform on a (much) larger space. Consider the topological dual $\mathscr{S}'(\mathbb{R}^n)$ of $\mathscr{S}(\mathbb{R}^n)$. This is the space of tempered distributions. We can identify every locally integrable function $f\colon \mathbb{R}^n \to \mathbb{C}$ of at most polynomial growth, or every $L^p$-function (an $L^p$-function need not have polynomial growth, but is nevertheless well-behaved enough, large growth can occur only on small sets due to the global integrability condition), with a tempered distribution by defining

$$T_f \colon \varphi \mapsto \int_{\mathbb{R}^n} f(x)\cdot \varphi(x)\,dx,$$

thus we see that $\mathscr{S}'(\mathbb{R}^n)$ is a large space, and in particular it contains $x \mapsto \lVert x\rVert$.

Since $\mathscr{F}_s$ is an automorphism, its transpose (or dual, or adjoint) $\mathscr{F}_s' \colon \mathscr{S}'(\mathbb{R}^n) \to \mathscr{S}'(\mathbb{R}^n)$ is also an automorphism. From $(3)$ we obtain

$$T_{\mathscr{F}_s[\varphi]}(\psi) = \int_{\mathbb{R}^n} \mathscr{F}_s[\varphi](x)\psi(x)\,dx = \int_{\mathbb{R}^n} \varphi(y)\mathscr{F}_s[\psi](y)\,dy = \bigl(\mathscr{F}_s'(T_{\varphi})\bigr)(\psi),$$

and thus it makes sense to call

$$\mathscr{F}_t := \mathscr{F}_s'$$

the Fourier transform of tempered distributions.

Similarly to the Fourier transform, we can extend differentiation to tempered distributions. If a tempered distribution is induced by a sufficiently often classically differentiable function with suitable growth restrictions, we can integrate by parts to obtain

$$\int_{\mathbb{R}^n} \bigl(D^{\alpha} f\bigr)(x)\varphi(x)\,dx = (-1)^{\lvert\alpha\rvert}\int_{\mathbb{R}^n} f(x)\bigl(D^{\alpha}\varphi\bigr)(x)\,dx,$$

and thus defining

$$D^{\alpha}T \colon \varphi \mapsto (-1)^{\lvert\alpha\rvert}\cdot T(D^{\alpha}\varphi)$$

extends the classical notion of differentiation to tempered distributions (we use the fact that $D^{\alpha} \colon \mathscr{S}(\mathbb{R}^n) \to \mathscr{S}(\mathbb{R}^n)$ is continuous).

In this setting, the Fourier transform of $f\colon x \mapsto \lVert x\rVert$ is defined (as the Fourier transform of the tempered distribution induced by $f$), and the formula for the derivatives (note that for $n > 1$, we can't write $f'$, we need to consider partial derivatives)

$$\mathscr{F}_t[D^{\alpha} f](\omega) = i^{\lvert\alpha\rvert}\omega^{\alpha}\mathscr{F}[f](\omega)$$

holds - where $D^{\alpha} f$ has to be interpreted as the distributional derivative in general. The function $f(x) = \lVert x\rVert$ is classically differentiable on $\mathbb{R}^n\setminus \{0\}$ (assuming we speak of the Euclidean norm), and thus we can also look at the function $D^{\alpha} f$ defined on $\mathbb{R}^n\setminus\{0\}$ by the classical derivative. For $\lvert\alpha\rvert$ small enough, this is compatible with the distributional derivative. For $\lvert\alpha\rvert$ too large, the classical derivative on $\mathbb{R}^n\setminus \{0\}$ doesn't induce the distributional derivative. For example in the case $n = 1$, we have $f'(x) = \operatorname{signum} x$ and the second distributional derivative comes out as $2\cdot \delta$ (with the Dirac distribution $\delta$), whereas the second classical derivative on $\mathbb{R}\setminus \{0\}$ vanishes identically.

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