Consider a standard setting for the development of the theory of distributions. Let $D(\Omega)$ be the space of test functions and $D'(\Omega)$ be the space of distributions ("generalized functions").

Can $\langle f,g\rangle$ be given a sensible definition for all $f,g\in D'(\Omega)$? Of course, $\langle f,\phi\rangle$ and $\langle \phi, f\rangle$ are well-defined and in $\mathbb{R}$ or $\mathbb{C}$ for any $\phi\in D(\Omega)$ and $f\in D'(\Omega)$. But more generally?

If a sensible definition can be made, it seems $\langle f,g\rangle$ may not always be a complex number. For example, for any $x\in\mathbb{R}$, there are test functions $\phi_n$ with $\phi_n\to\delta$ such that $\langle \delta, \phi_n\rangle=x$ for all $n$. Thus, $\langle \delta,\delta\rangle$ couldn't be a simple real number. But maybe a distribution?

My main motivation in asking this is to make sense of the following formula: $$ \langle \frac{1}{2\pi} e^{ikx}, e^{-ikx}\rangle = \delta. $$ This formula seems at first glance to be nonsense, but at the same time, we know its intended meaning is the Fourier integral formula: $$ f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x') e^{-ikx'} \ dx' \ e^{ik x} \ dk $$ for a wide class of functions $f:\mathbb{R}\to\mathbb{R}$.

I'm reading Strichartz' A Guide to Distribution Theory and Fourier Transforms. Maybe he will cover this at some point?

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    $\begingroup$ No it doesn't really make sense. The inner product notation is merely suggestive. It is just a way to pair an element with something in its dual space. What you have proposed does not obey this. $\endgroup$ Jun 18, 2014 at 5:30
  • $\begingroup$ $\langle \frac{1}{2\pi} e^{ikx}, e^{-ik'x}\rangle = \delta(k-k')$ - you shouldn't set $k'=k$ $\endgroup$
    – user8268
    Jun 18, 2014 at 7:04

1 Answer 1


A distribution is a linear operator $D(\Omega)\to\Bbb R $, so the product of distributions has no sense.

You can introduce convolution of distributions - however, you will need to impose some restrictions on the support of one of the distributions (moreover, you will lose associativity of convolution for general case).


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