A few simple "division" problems

Let $f(x)$ be a smooth function and $g$ be a distribution. How do we solve equations of the form
$f(x)T = g$ for $T$ ? If $T$ and $g$ were functions, it's all very easy. We can just divide by $f$ and we're happy, being just a bit careful whenever $f(x) = 0$ . In the world of distributions, this turns out to be horrible.

There is very little to no existing literature on this problem, which is a major motivation for this post. In this section, I'll discuss some easy cases which turn out to be sufficient for a respectable number of situations where distribution theory can be applied. Trying to do more than that is horrifyingly difficult and so a discussion of this will be postponed to the very end of this post.

Basics

Let's stay in one dimension and take $f(x) = x$ as an example, so that the problem of interest is finding all $T$ satisfying $xT = 0$ . If $x \neq 0$ , intuitively it may make sense to "divide" by $x$ to find $T = 0$ . But at $x=0$ we have a problem. So, again intuitively, $T$ could have some "weight" at $0$ . We thus predict that solutions to the equation
$x \cdot T = 0$ in the sense of distributions are $T = a\delta_0$ , where $\delta_0$ is the Dirac delta and $a$ is a constant.

This is correct. Here is how one proves this.

Proof

How about if the RHS isn't zero? If we have something like $xT = f$ , for a smooth function $f$ , we can write $f = f \cdot 1/x \cdot x$ here $1/x$ is the principal value. Now we can rearrange to $x(T - f \cdot 1/x) = 0$ and apply the result we have for solving $xT = 0$ . Sadly this doesn't work so well when $f$ is distribution, since then $\left(\operatorname{p.v.}\frac1x\right)f$ would be a product of distributions, which is (in general) illegal.

Of course, to make any of that work you need to justify this little detail: That $x \cdot \operatorname{p.v.}\frac1x = 1$ . This is pretty easy by chasing the definitions so I'll leave that to you.

Exercise: Solve $xT = S$ for $S$ a given distribution. (I will provide the solution in the final stretch of this post.)

In sum, we've learned how to "undo multiplication by $x$ " in certain situations. To generalize, it's best to introduce a the notion of the support of a distribution. (i tried to work it out without this and i couldnt figure it out. i'm including more detail here compared to the rest of this post because i think current literature on this is not very well-saturated.)

Definition (Support): Let $T$ be a distribution.

$T$ is said to vanish on an open set $U$ if $\langle T,\phi\rangle = 0$ for all $\phi$ supported in $U$ .
Where $T$ "vanishes" is then the union of all such $U$ .
The support of $T$ is the complement of where $T$ vanishes.

For this section, we're primarily interested in the case where the support of $T$ is extremely degenerate --- something like a single point or a finite set. This is relevant because of the following.

Proposition: Suppose $fT = 0$ . Then the support of $T$ is contained in $\{f=0\}$ .

For example, if $f$ is a polynomial, then $T$ can only be supported at the roots of $f$ . Before, we saw the case when $f = x$ , in which case, $T$ 's support must thus be contained in $\{0\}$ .

Proof of Proposition

Warning: The converse to this proposition does NOT hold! Try to guess an example.

Example

Now we reach the point of all this.

Theorem: In arbitrary dimensions: Suppose the support of $T$ is $\{a\}$ . Then $T = D\delta_a$ where $\delta_a$ is the Dirac delta at $x=a$ , and $D$ is some linear differential operator. (This is also an iff, by the way --- the converse isn't hard to show)

By "some linear differential operator" I mean that $D = \sum_{|\alpha| \leq j} c_\alpha\partial^\alpha$ for some order $j < \infty$ and some coefficients $\{c_\alpha\}$ .

Just as an aside, this theorem appears to suggest that any such distribution $T$ must have finite order (given by the order of $D$ ). This is true, and in fact, any $T$ whose support is compact will have finite order. (exercise!)

A proof of this theorem can be found in Rudin's Functional Analysis.

One particularly useful "instance" of this result is the following.

Proposition: Stay in 1D. Fix $a \in \mathbb{R}$ . We have that $T$ satisfies $(x-a)^nT = 0$ iff $T = \sum_{k=0}^{n-1}c_k\delta_a^{(k)}$ for some constants $c_0,\dots,c_{n-1}$ . More succinctly, $T = D\delta_a$ where $D$ is a linear differential operator of order at most $n-1$ .

For example, the solutions to $x^2T = 0$ are $T = a\delta_0 + b\delta_0'$ for some $a$ and $b$ .

Proof

Note that it's rather awkward to try generalizing this proposition to higher dimensions. Some examples of attempts that fail:

It's tempting to conjecture that solutions to $|x|^nT$ are of the form $D\delta_0$ for $D$ a linear differential operator of order less than $n$ . The issue is that $|x|^n$ is not smooth for odd $n$ , so the problem is ill-posed.
So maybe we can try looking instead at smooth polynomials that vanish at the origin to some order. For example $xy$ vanishes to 2nd order at $(0,0)$ . So perhaps we'd like to study solving $xyT = 0$ . However this is quite hard because $xy$ vanishes at many more points than just $(0,0)$ , so the support of $T$ can be quite wild.

So, sticking in one dimension, here is a more general theorem. Its proof is optional as the theorem afterwards is strictly better and is proved in a completely different manner.

Theorem 6: Let $f:\mathbb{R} \to \mathbb{R}$ (or $\to \mathbb{C}$ ) be a smooth function with $\{x : f(x) = 0\} = \{x_1,x_2,\dots,x_n\}$ . Suppose that for each $j$ , the order of the zero $x=x_j$ is $k_j < \infty$ in the sense that $\lim_{x \to x_j} \frac{f(x)}{(x-x_j)^{k_j}}$ exists and is non-zero. (Equivalently, $k_j$ is the least positive integer for which $f^{(k_j)}(x_j) \neq 0$ .) Then the solution to $fT = 0$ is given by
$T = \sum_{j=1}^n D_j\delta_{x_j}$ where $D_j$ is a linear differentiation operator of order at most $k_j-1$ . Here, $\delta_{x_j}$ is the Dirac delta at $x_j$ .

The specific case $f(x) = (x-a)(x-b)$ , $a \neq b$ is enlightening and of independent interest, so I'll first present the proof for this instance as it looks nicer.

Proof

Now, the proof of the theorem.
Full Proof

Theorem 6 is already pretty good for several of the applications I will present near the end. If you want more than that, we need to start getting tricky. Let's go one more step up (for good reason, I promise!).

Theorem 6.5: Let $f:\mathbb{R} \to \mathbb{R}$ (or $\to \mathbb{C}$ ) be a smooth function with $\{x : f(x) = 0\}$ is an at-most countable set $Z$ . Suppose that for each $z \in Z$ , the order of the zero $x=z$ is $k_z < \infty$ in the sense that $\lim_{x \to z} \frac{f(x)}{(x-z)^{k_z}}$ exists and is non-zero. (Equivalently, $k_z$ is the least positive integer for which $f^{(k_z)}(z) \neq 0$ .) Then the solution to $fT = 0$ is given by
$T = \sum_{z \in Z} D_z\delta_{z}$ where $D_z$ is a linear differentiation operator of order at most $k_z-1$ . Here, $\delta_{z}$ is the Dirac delta at $z$ .

If $\sum_{z \in Z}$ makes you nervous: It turns out that if all zeroes are of finite order then $Z$ actually must be quite nice, in that it cannot have any accumulation points! There is a very clever way to argue this quickly: Proof

Now let's prove Theorem 6.5. With infinitely many zeroes, the proof for Theorem 6 does not work. We're going to do something completely different.

Proof

Alright that's it for now. Going any further will be venturing into diabolical territory.

Further operations with distributions

Fourier Transform

I've been teasing this for a while --- finally we come to the point of introducing tempered distributions. Recall that $\mathcal{S}(\mathbb{R}^N)$ is the space of Schwartz functions --- a larger space than the test functions for usual distributions.

There are countless different definitions for the Fourier transform, depending on how you prefer certain applications of it to be normalized. Here is the definition I'll be working with.

Definition: The Fourier transform is the linear map $\mathcal{F}: \mathcal{S}(\mathbb{R}^N) \to \mathcal{S}(\mathbb{R}^N)$ given by
$(\mathcal{F}(u))(\xi) := \int_{\mathbb{R}^N} u(x)e^{-2\pi i \xi \cdot x}\,dx. \qquad (*)$ The Fourier transform is also denoted as $\hat{u}$ .

Of course, one should prove that $\mathcal{F}$ actually sends $\mathcal{S}(\mathbb{R}^N)$ to $\mathcal{S}(\mathbb{R}^N)$ , but this is well-documented so I will start omitting proofs again. Instead, here are the key properties (independent of distribution theory).

$\mathcal{F}$ is actually continuous (with respect to the topology on $\mathcal{S}(\mathbb{R}^N)$ ). One way to make this explicit is as follows: If $f_k \to f$ in $\mathcal{S}(\mathbb{R}^N)$ (that is, $\|f-f_k\|_{m,n} \to 0$ for all $m$ and $n$ ) then $\hat{f_k} \to \hat{f}$ .
Using the formula $(*)$ , we can define $\mathcal{F}$ over $L^1(\mathbb{R}^N)$ , in which case
$\mathcal{F}:L^1 \to L^\infty$ .
Similarly, for $1 \leq p < 2$ , $\mathcal{F}$ can be viewed as a map
$\mathcal{F}:L^p \to L^{p'}$ where $p'$ is the Holder conjugate of $p$ . This is because for all such $p$ , the integral in $(*)$ is well-defined for $u \in L^p$ .
But for $p=2$ , $(*)$ is nonsense. Nevertheless, one can extend $\mathcal{F}$ to a map $L^2 \to L^2$ by using a density argument.
Don't bother taking the Fourier transform of a function in $L^p$ for $p > 2$ . (unless tempered distributions can save us...?)
Thus the Fourier transform only reasonably sends $u$ to a function if $u \in L^p$ for $1 \leq p \leq 2$ .
The inverse Fourier transform $\mathcal{F}^{-1}(u)$ , denoted also as $\check{u}$ , is given by
$\mathcal{F}^{-1}(u)(x) = \int_{\mathbb{R}^N} u(\xi)e^{2\pi i x \cdot \xi}\,d\xi.$ This satisfies the same qualitative properties as the Fourier transform.

This is a true inverse of the Fourier transform over $\mathcal{S}(\mathbb{R}^N)$ . When $u$ lives in a larger space like $L^1$ , things get far dicier since $\mathcal{F}^{-1}$ would have to be a map $L^\infty \to L^{1}$ which we cannot expect it to be. In other words, we cannot expect the tempting identity
$u(x) \stackrel?= \int_{\mathbb{R}^N} \hat{u}(\xi)e^{2\pi i x \cdot \xi}\,d\xi$ to actually make sense when we assume only that $u \in L^1$ . So don't write this! Info blast
In harmonic analysis this motivates the question of whether or not we can write
$u(x) \stackrel?= \lim_{n \to \infty}\int_{E_n} \hat{u}(\xi)e^{2\pi i x \cdot \xi}\,d\xi$ for a suitable meaning of " $=$ " and a particular choice of sets $E_n$ that expand to $\mathbb{R}^N$ , such as $E_n = [-n,n]^N$ or $E_n = B(0,n)$ . For bizarre reasons, the former choice works fine for somewhat boring reasons, whereas the latter choice (which is arguably more natural) fails spectacularly due to Kakeya sets (!).

The moral of the story is that to do even the most basic and intuitive thing in Fourier analysis can result in surprisingly gnarly situations.
We have that $\hat{\overline{u}} = \overline{\check{u}}$ and $\check{\overline{u}} = \overline{\hat{u}}$ . ("The direction of the arrow flips when we move it across the bar.") This helps explain why $\mathcal{F}$ and $\mathcal{F}^{-1}$ have the same qualitative properties. For example, since Riemann-Lebesgue says that $u \in L^1 \implies \hat{u} \in C_0$ , we must therefore also have that $u \in L^1 \implies \check{u} \in C_0$ .
$e^{-\pi|x|^2}$ is invariant under the Fourier transform.
(Uncertainty principle) For $u \neq 0$ , $u$ and $\hat{u}$ cannot both be compactly supported.
(Riemann-Lebesgue) If $u \in L^1$ , then $\hat{u}$ is, in fact, continuous and decays to 0 at $\infty$ . (This is a major improvement over the previously-stated result $\mathcal{F}:L^1 \to L^\infty$ !)
(Integrability turns to smoothness) If $u$ is compactly supported then $\hat{u}$ is smooth. (The reverse is not true.)
(Fourier transform brutalizes derivatives) If $u \in \mathcal{S}(\mathbb{R}^N)$ then
$\widehat{\partial_{x_j}u}(\xi) = 2\pi i \xi_j\hat{u}(\xi).$ More generally
$\widehat{\partial^\alpha u}(\xi) = (2\pi i \xi)^\alpha\hat{u}(\xi)$ for any multi-index $\alpha$ . (The hypotheses on $u$ can be weakened but the exact requirements are not relevant for our situation.) This property shows why Fourier transforms can destroy PDEs.
(Convolution becomes multiplication) We have
$\widehat{u \star v} = \widehat{u}\widehat{v}.$
(Translation becomes a phase factor) Let $g(x) = f(x-a)$ . Then
$\hat{g}(\xi) = e^{-2\pi i \xi \cdot a} \hat{f}(\xi).$
(Plancherel) For any $u,v \in \mathcal{S}(\mathbb{R}^N)$ (or $L^2$ ) we have
$\int_{\mathbb{R}^N} \hat{u}v\,dx = \int_{\mathbb{R}^N} u\hat{v}\,dx.$ This can be written in any of the following alternative ways:
$\int_{\mathbb{R}^N} \hat{u}\check{v}\,dx = \int_{\mathbb{R}^N} uv\,dx$ $\int_{\mathbb{R}^N} \hat{u}\overline{v}\,dx = \int_{\mathbb{R}^N} u\overline{\check{v}}\,dx$ $\int_{\mathbb{R}^N} \hat{u}\overline{\hat{v}}\,dx = \int_{\mathbb{R}^N} u\overline{v}\,dx$ Since the inner product over $L^2(\mathbb{R}^N;\mathbb{C})$ is $(u,v) := \int_{\mathbb{R}^N} u\overline{v}\,dx$ , we therefore see that
$(u,v) = (\hat{u},\hat{v}).$ (see the next info blast for an explanation of why i'm using the round brackets to denote inner product)

This says that the Fourier transform is an isometry over $L^2$ . In particular, the Fourier transform preserves distances: By taking $u = v$ above we deduce Pareseval's identity,
$\int_{\mathbb{R}^N} |u|^2\,dx = \int_{\mathbb{R}^N} |\hat{u}|^2\,dx.$

Now let's bring tempered distributions into the mix. The motivation here is that requiring that heavy integrability and/or smoothness conditions on $u$ to ensure it plays well with the Fourier transform can be very annoying, as it prevents us from taking the Fourier transform of various functions that we'd like to apply the advantages of Fourier analysis to. Viewing everything in terms of tempered distributions fixes all this.

Recall that a tempered distribution $T$ is a continuous linear functional on the Schwartz functions $\mathcal{S}(\mathbb{R}^N)$ . Recall also that any function in $L^p$ corresponds to a tempered distribution (and can thus be viewed as a tempered distribution). Motivated by Plancherel's theorem,
$\int \hat{u}v = \int u\hat{v},$ we see that for all functions $u$ and $v$ we can say that $\langle \hat{u},v\rangle = \langle u,\hat{v}$ , where we recall that by $\langle f,g\rangle$ for functions $f,g$ we just mean $\int fg$ . Therefore, the most natural way to define the Fourier transform for a tempered distribution is as follows.

Definition: Let $T$ be a tempered distribution. Then $\hat{T}$ is a tempered distribution defined as
$\langle \hat{T}, \phi\rangle = \langle T, \hat{\phi}\rangle.$ It's quite simple! You just move the hat over. Info blast

Notes:

As usual, it needs to be proven that $\hat{T}$ is indeed a tempered distribution.
Note that this definition does NOT work when $T$ is only a distribution. This is because if $\phi$ were only a $C_c^\infty$ test function, there is no guarantee that $\hat{\phi}$ will still live in $C_c^\infty$ , meaning that $\langle T,\hat{\phi}\rangle$ won't be defined. (IN FACT --- it's actually completely impossible for $\hat{\phi} \in C_c^\infty$ by the uncertainty principle!) This is why the Schwartz functions, being a space over which the Fourier transform is a bijection, is far more ideal.
This clearly agrees with the usual notion of Fourier transform when $T$ is actually a (Schwartz) function.

A few quick examples to get our feet wet.

We have that $\hat{\delta_0} = 1$ . Indeed,
$\langle \hat{\delta_0},\phi\rangle = \langle \delta_0,\hat{\phi}\rangle = \hat{\phi}(0)$ $= \int_{\mathbb{R}^N} \phi(x)e^{-2\pi i(0 \cdot x)}\,dx = \int_{\mathbb{R}^N} \phi(x) \cdot 1\,dx = \langle 1, \phi\rangle.$ Similarly, $\hat{1} = \delta_0$ .
More generally, we have that $\hat{\delta_a} = e^{2\pi \xi \cdot a}$ and $\widehat{e^{2\pi x \cdot a}} = \delta_a$ . This can also be seen by applying the translation property of Fourier transforms, which generalizes to tempered distributions.

Properties

Unsurprisingly many of the properties for the Fourier transform over functions will carry over to tempered distributions. For instance:

Fourier transform is linear.
We have $\widehat{\partial^\alpha T} = (2\pi i \xi)^\alpha \hat{T}$ .
If we define $\langle \check{T},\phi\rangle := \langle T, \check{\phi}\rangle$ then $\hat{\check{T}} = T$ .

You can similarly obtain analogues for the "translation property" of Fourier transforms, among other such algebraic properties. Properties such as Parseval's identity don't have a great analogue in distributions.

There's one particularly remarkable property which unexpectedly carries over, however!

Theorem (Schwartz-Paley-Wiener): Suppose $T$ is a tempered distribution with compact support. Then $\hat{T}$ is a smooth function!

Notes:

This is an extremely powerful generalization of the fact that the Fourier transform of a compactly supported function is a smooth function. This is not too hard to prove using integration by parts and whatnot, but this becomes quite a bit more challenging to prove for tempered distributions.
In fact it turns out that $\hat{T}$ will be an entire function, i.e. analytic when extended to a function on $\mathbb{C}^N$ .
A quick corollary is that if $\hat{T}$ has compact support then $T$ is a smooth function.
As an example, $\delta_0$ has compact support. And indeed, $\hat{\delta_0} = 1$ which is smooth.

Fourier Transform of measures

Measures are distributions, so we can Fourier them (provided they are tempered)! For a tempered measure $\mu$ , we have
$\langle \hat{\mu},\phi\rangle = \langle \mu, \hat{\phi}\rangle = \int_{\mathbb{R}^N} \hat{\phi}(\xi)\,d\mu(\xi).$ In general that's all we can say. But if $\mu$ is a finite measure we can write a bit more, using Fubini's theorem to write
$= \int_{\mathbb{R}^N} \int_{\mathbb{R}^N} e^{-2\pi i \xi \cdot x}\phi(x)\,dx\,d\mu(\xi)$ $= \int_{\mathbb{R}^N} \int_{\mathbb{R}^N} e^{-2\pi i \xi \cdot x}\phi(x)\,d\mu(\xi)\,dx$ $= \left\langle \int_{\mathbb{R}^N} e^{-2\pi i \xi \cdot x}\,d\mu(\xi), \phi(x)\right\rangle.$ So we deduce the following.

Proposition: If $\mu$ is a finite measure then its Fourier transform is a continuous function.

Proof of continuity

We can see this as a generalization of Riemann-Lebesgue to distributions, except we lose the additional property that the Fourier transform decays to $0$ . For example, $\hat{\delta_0} = 1$ is continuous but certainly does not decay to $0$ at infinity.

A few more examples

We can Fourier cosine and sine. Since $\cos(2\pi x) = \frac{e^{2\pi i x} + e^{-2\pi i x}}{2}$ , it follows that $\widehat{\cos(2\pi x)} = \frac12\delta_1 + \frac12\delta_{-1}$ . Similarly, $\widehat{\sin(2\pi x)} = \frac1{2i}\delta_1 - \frac1{2i}\delta_{-1}$ .
Let $d\omega$ be the surface measure on the unit ball, then $\widehat{d\omega}$ exists. But $d\omega$ is supported in the unit ball, hence is compactly supported, so in fact $\widehat{d\omega}$ is a smooth function given by
$\widehat{d\omega}(\xi) = \int_{\mathbb{R}^N} e^{-2\pi i \xi \cdot x}\,d\omega = \int_{\partial B(0,1)} e^{-2\pi i \xi \cdot x}\,d\mathcal{H}^{N-1}$ where $\mathcal{H}^{N-1}$ denotes Hausdorff measure (this is just a surface integral). Using the method of stationary phase, one can find the decay estimate
$|\widehat{d\omega}(\xi)| \leq \frac{C}{|\xi|^{\frac{N-1}{2}}},$ which is important in harmonic analysis.
$\arctan x$ is not integrable, so its Fourier transform cannot be computed using the integral definition. Nevertheless, one may attempt the following symbolic computation:
$\widehat{\arctan}(\xi) = \frac{1}{2\pi i \xi} \cdot 2\pi i \xi\widehat{\arctan}(\xi) = \frac{1}{2\pi i \xi}\widehat{\frac{d}{dx}\arctan x}$ $= \frac{1}{2 \pi i \xi}\widehat{\frac{1}{1+x^2}}$ Now, $\frac{1}{1+x^2}$ can be Fourier'd in the classical sense, and by some complex analysis one can compute that $\widehat{\frac{1}{1+x^2}}(\xi) = \pi e^{-2\pi|\xi|^2}$ . So it is very tempting to conclude that
$\widehat{\arctan}(\xi) \stackrel?= \frac{1}{2i\xi}e^{-2\pi|\xi|^2}.$ It turns out this is mostly true. To make this rigorous, we first regard $\arctan$ as a tempered distribution (it's bounded!), which makes the computation
$2\pi i \xi \cdot \widehat{\arctan}(\xi) =\widehat{\frac{d}{dx}\arctan x} = \widehat{\frac{1}{1+x^2}} = \pi e^{-2\pi|\xi|^2}$ perfectly rigorous. (Note that multiplying by $\frac{1}{2\pi i \xi}$ is illegal as it is not smooth.) But now what? To make the division by $2\pi i \xi$ rigorous, we write
$\pi e^{-2\pi|\xi|^2} = 2 \pi i \xi \cdot \frac{1}{2i} \cdot \operatorname{p.v.}\frac{e^{-2\pi|\xi|^2}}{\xi}$ Now
$2 \pi i \xi\left(\widehat{\arctan}(\xi) - \frac{1}{2i} \cdot \operatorname{p.v.}\frac{e^{-2\pi|\xi|^2}}{\xi}\right) = 0$ so that $\widehat{\arctan}(\xi) - \frac{1}{2i} \cdot \operatorname{p.v.}\frac{e^{-2\pi|\xi|^2}}{\xi} = a\delta_0$ for some constant $a$ . To quickly find $a$ , test each side against $e^{-\pi \xi^2}$ , which is invariant under Fourier transform. We find that
$\langle \widehat{\arctan}(\xi), e^{-\pi \xi^2}\rangle = \langle \arctan x, e^{-\pi x^2}\rangle = 0$ and that
$\left\langle \operatorname{p.v.}\frac{e^{-2\pi|\xi|^2}}{\xi},e^{-\pi \xi^2}\right\rangle = \lim_{\varepsilon \to 0^+} \int_{\mathbb{R} \setminus (-\varepsilon,\varepsilon)} \frac{e^{-2\pi|\xi|^2}}{\xi}e^{-\pi \xi^2}\,d\xi = 0,$ whereas $\langle a\delta_0, e^{-\pi \xi^2}\rangle = a$ . Thus $0 - 0 = a$ which gives $a = 0$ .