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Let $l_1$ and $l_2$ be two distributions in disjoint variables $x_1, ..., x_n$ and $y_1, ..., y_m$. Then it is said to be possible to define a product distribution.

However, I am fundamentally confused. Distributions are in fact linear functionals on the space of smooth and compactly supported functions. Then, how does the 'product' of linear functionals again become a linear functional?

Especially, what can be a definition of $\delta(x_1)\delta(x_2)$ such that it is equal to $\delta(x_1, x_2)$? I am just stuck......

Keith
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3 Answers3

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As mathcounterexamples.net says, this is related to the tensor product of distributions. Here is an outline of how to do this. The full proof is rather involved, depending on your background knowledge, so I'll give a reference.

Let's write $\newcommand{\D}{\mathcal{D}}\newcommand{\R}{\mathbb{R}}\D(\R^n)$ for the space of smooth, compactly supported real-valued functions on $\R^n$. We write $\D'(\R^n)$ for the dual space, the space of distributions (continuous linear functions $\D(\R^n) \rightarrow \R$). So your $l_1 \in \D'(\R^n)$ and $l_2 \in \D'(\R^m)$. To avoid confusion with the pointwise product, I'm going to write the product in $\D'(\R^{n+m})$ as $l_1 \otimes l_2$. But in order to define this, I'll define the analogous operation for $\D(\R^{n+m})$ first.

If $f \in \D(\R^n)$ and $g \in \D(\R^m)$, define $f \otimes g \in \D(R^{n+m})$ by: $$ (f \otimes g)(x_1, \ldots, x_n, y_1, \ldots, y_m) = f(x_1,\ldots,x_n)g(y_1,\ldots,y_m) $$ Importantly, the linear span of $\{ f \otimes g \mid f \in \D(\R^n) \text{ and } g \in \D(\R^m) \}$ is dense in $\D(\R^{n+m})$ its usual topology. This means we can uniquely define distributions in $\D'(\R^{n+m})$ by defining them on smooth functions of the form $f \otimes g$.

So for $l_1 \in \D'(\R^n)$ and $l_2 \in \D'(\R^m)$, define $$ (l_1 \otimes l_2)(f \otimes g) = l_1(f) l_2(g), $$ or more generally $$ (l_1 \otimes l_2)\left(\sum_{i=1}^p \alpha_i f_i \otimes g_i\right) = \sum_{i=1}^p \alpha_i l_1(f_i) l_2(g_i), $$ where $\alpha_i \in \R$, $f_i \in \D(\R^n)$ and $g_i \in \D(\R^m)$. This defines a continuous linear map, which can therefore be extended to $l_1 \otimes l_2 : \D(\R^{n+m}) \rightarrow \R$, defining a distribution.

To fill in the gaps in what I'm saying above, consult a standard reference text on distributions, such as Hörmander's The Analysis of Linear Partial Differential Operators I, Theorem 5.1.1.

You asked to see how this works out for $\delta$ functions. So for any $h \in \D(\R^p)$, $\delta(h) = h(0)$ is the usual definition. Now, two continuous linear maps that agree on a subset of $\D(\R^{n+m})$ whose linear span is dense must be equal. To avoid confusion, let's write $\delta_n$ for the $\delta$ function in $\D'(R^n)$. So we see that $$ (\delta_n \otimes \delta_m)(f \otimes g) = \delta(f)\delta(g) = f(0)g(0) = (f \otimes g)(0) = \delta_{n+m}(f \otimes g), $$ and this proves that $\delta_n \otimes \delta_m = \delta_{n+m}$.

Robert Furber
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Concretely for $\phi \in C^\infty_c(\Bbb{R}^2)$ take $\varphi_n,\psi_n \in C^\infty_c(\Bbb{R})$ such that $$\sum_{n=1}^N \varphi_n(x)\psi_n(y)\to \phi$$ in test function topology.

Then for $T,S\in D'(\Bbb{R})$ $$\langle T(x)S(y),\phi\rangle =\sum_{n=1}^\infty \langle T(x)S(y),\varphi_n(x)\psi_n(y)\rangle=\sum_{n=1}^\infty \langle T,\varphi_n\rangle \langle S,\psi_n\rangle$$

To find $\varphi_n,\psi_n$, for $\phi$ supported on $(-T,T)\times (-T,T)$, take $\rho \in C^\infty_c(\Bbb{R})$, $\rho = 1$ on $(-T,T)$ and look at the Fourier series $$\phi(x,y) = \sum_{m,k}c_{m,k}e^{2i\pi mx/T}\rho(x)e^{2i \pi ky/T}\rho(y)$$

reuns
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  • Yes, this is one way of proving that $\mathcal{D}(\mathbb{R}^n) \otimes \mathcal{D}(\mathbb{R}^m)$ is dense in $\mathcal{D}(\mathbb{R}^{n+m})$. – Robert Furber Sep 26 '19 at 02:01
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If you want to multiply $\delta(x_1)$ and $\delta(x_2)$ you first need to make them into functions acting on the space space, so you multiply $\delta(x_1)Id(x_2)$ and $Id(x_1)\delta(x_2)$, where $Id$ is just the identity map. With that interpretation you get $\delta(x_1)\delta(x_1)=\delta(x_1, x_2)$.

quarague
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