How can tempered distributions be identified with functions?

Question

In Stein's Harmonic Analysis: Real-Variable Methods, Orthogonality and Oscillatory Integrals, he defines tempered distribution ($\mathscr S'$) as continuous linear functionals from the Schwartz class. Here, the continuity is given with respect to the family of seminorms \begin{equation*}\|\Phi\|_{\alpha,\beta}=\sup_{x\in\mathbb R^N}|x^\alpha\partial^\beta_x\Phi(x)|.\end{equation*} Then, without any further clarification, he proceeds to discuss convolutions between a tempered distribution and functions in the Schwartz family. This is where I get puzzled. If the tempered distributions are functionals, what is this convolution supposed to mean? It seems as if he (and every other source, for that matter) was assuming these functionals can be clearly identified with some functions. My question is: how do you make this identification?

I was thinking, as Schwartz functions are in $L^2$, maybe the identification is the one given by Riesz representation theorem. However, I think this is not possible as the topology we are considering in the Schwartz class is different from that of $L^2$. Moreover, while discussing $H^p$ spaces, he claims that, for $p>1$, $L^p$ is the same as $H^p$. Here, he is using again this identification that I don't quite get and, if my hypothesis was correct that the identification is made through Riesz representation theorem, this should mean $H^2=L^2=\mathscr S'$. This seems a bit strange to me. Another thing that's worrying me as well is the fact that he is discussing, without a prior definition, bounded distributions. Of course, if these were elements of $L^2$'s dual, they would be automatically bounded, so this is another hint that my original assumption about the identification is wrong.

I think this is a very basic question, but I don't find any source in which this is specifically discussed and clarify. How can we talk about an object as both a tempered distribution (an element of $\mathscr S'$) and a function defined on $\mathbb R^N$?

Answer 1

We don’t need to identify distributions with functions to meaningfully discuss convolution with distributions. Most distributions can’t be identified with functions. Any operation normally defined between functions we define for distributions using duality, by determining how it acts on Schwartz functions. Let $\langle g,\phi\rangle$ be the evaluation pairing between a distribution $g$ and a Schwartz function $\phi$, i.e. $g: \phi \mapsto \langle g,\phi\rangle$. Notice that if $g$ is a locally integrable function, this pairing is just the $L^2$ inner product.

The strategy for determining the proper dual formulation for a statement about distributions is to assume all objects involved are nice functions, then move things around until we have an evaluation pairing between a distribution and a Schwartz function. For example, to make sense of the the action of the distribution $g*\psi$ on a Schwartz function $\phi$, where $g$ is a distribution and $\psi$ is Schwartz, if $g$ were nice we could use Fubini to get

$$\begin{align} \langle g*\psi, \phi \rangle &= \int \phi(x) \int g(y)\psi(x-y) dydx \\ &= \int g(y) \int \phi(x)\psi(x-y)dx dy \\ &= \langle g, \phi*R\psi\rangle \end{align}$$

Where $R$ is the reflection operator $R:\psi(x)\mapsto \psi(-x)$. With this in mind, we define by duality $\langle g*\psi, \phi \rangle = \langle g, \phi*R\psi\rangle$. This determines the behavior of the distribution $g*\psi$ uniquely, since we know how $g$ acts on Schwartz functions. Of course, there’s some details here you can check, like the fact that $g*\psi$ is indeed a tempered distribution, that convolution of Schwartz functions is Schwartz, etc.

Another example: to define derivatives of a distribution, we determine its action on smooth functions by pretending the distribution is smooth and integrating by parts until we have something meaningful:

$$ \langle \partial^\alpha g,\phi\rangle = (-1)^{|\alpha|} \langle g, \partial^\alpha \phi\rangle$$

This equality ultimately defines $\partial^\alpha g$.

Once you’ve established these duality-based definitions, you can informally treat distributions as if they’re functions, even though to be precise you need to work with their actions on Schwartz functions. This is what’s going on behind the scenes in Stein.

Answer 2

A distribution is defined as a linear functional over a space of test functions but can (should?) be thought of almost as functions on $\mathbb R^N.$ That's because distributions have local behavior and identity.

If $u$ is a distribution and $\varphi$ a test function then a common way to write the action of $u$ on $\varphi$ is $\langle u, \varphi \rangle.$ This is because there are similarities with inner products. For example, the action of an ordinary function $f$ on a test function $\varphi$ is $$ \langle f, \varphi \rangle = \int_{\mathbb R^N} f(x) \, \varphi(x) \, d^Nx. $$

A distribution can be translated. The definition is defined based on how it works for ordinary functions: $$ \langle \tau_a f, \varphi \rangle = \int (\tau_a f)(x) \, \varphi(x) \, d^Nx = \int f(x-a) \, \varphi(x) \, d^Nx = \int f(x) \, \varphi(x+a) \, d^Nx = \langle f, \tau_{-a}\varphi \rangle . $$ We therefore define the translation $\tau_a u$ by $\langle \tau_a u, \varphi \rangle = \langle u, \tau_{-a}\varphi \rangle$ for all test functions $\varphi.$

I'm not sure that this answers your questions, but hopefully it helps a little bit in understanding distributions.

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EDIT: Convolution of two distributions

If $f$ and $g$ are ordinary functions for which the convolution $f*g$ is defined then $$ \langle f*g, \varphi \rangle = \int (f*g)(x) \, \varphi(x) \, dx = \int \left( \int f(x-y) \, g(y) \, dy \right) \varphi(x) \, dx \\ = \iint f(x-y) \, g(y) \, \varphi(x) \, dy \, dx = \iint f(z) \, g(y) \, \varphi(z+y) \, dy \, dz \\ = \langle (f\otimes g)(z,y), \varphi(z+y) \rangle = \langle f\otimes g, \varphi\circ+ \rangle. $$

Therefore the convolution of two distributions is defined by $$ \langle u*v, \varphi \rangle = \langle u\otimes v, \varphi\circ+ \rangle, $$ where the tensor product of two distributions is defined by $$ \langle (u\times v)(x,y), \varphi(x,y) \rangle = \langle u(x), \langle v(y), \varphi(x,y) \rangle\rangle. $$ Here the variables $x$ and $y$ are formal, especially after $u$ and $v$, used to make it clear how things are paired.

Answer 3

The idea would be to define the tempered distribution $T_{\phi}$ for some Schwartz function $\phi$ analogously to what is done for test functions by $$T_{\phi}(\psi)=\int \phi\psi \,d\lambda.$$ This implies that certain tempered distributions can be represented in integral form by Schwartz functions but in contrast to the Hilbert space $L^2$ not all elements in the dual of the Schwartz functions have such a representation (the norm plays an important role here since we are looking at the Banach space $\mathcal{S}$.)

Since test functions are dense in the Schwarz functions, most arguments which you can make on test functions translate directly to Schwartz functions as well. But it turns put that the dual of test functions is not the "correct" way of looking at distributions when it comes to generalizing the Fourier transform.

The convolution of a distribution $T$ with $f$ in the underlying Banach space is then defined as $$(T\ast f)(x):=T(f(x-\cdot)).$$ From the notion above you can in this way convolute tempered distributions and Schwartz functions.