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The unit doublet is a symbolic object whose convolution with a differentiable function is supposed to give the derivative:

$$(x * u_1)(t) = \frac{dx(t)}{dt}$$

See also:

Can this equation be made rigorous, in a way that it would be valid for all $x\in C^1(\mathbb R)$? I.e., is there a definition of the unit doublet as a distribution or a measure, and a version of the convolution operation that would make sense for all $C^1(\mathbb R)$ functions?

This is related to my question on MO: , but the theorems referred in the answers to that question include only a small fraction of $C^1(\mathbb R)$.

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up vote 3 down vote accepted

For $C^1(\mathbb{R})$ it is a bit delicate. But for $C^1(\mathbb{R})$ with compact support (or rapid decay a la Schwartz functions), you can define $u_1$ as (up to a negative sign) the distributional derivative of the Dirac distribution. It is a distribution of compact support, and so also a tempered distribution.

Using the fact that it has compact support, you should be able to convolve it against any distribution (see Hormander, Analysis of Linear Partial Differential Operator, Volume 1, Chapter 4). A function in $C^1$ is locally integrable, and hence a distribution. So the object $x*u_1$ is well defined as a distribution.

So at least distributionally the equation you wrote make sense: that acting on any smooth function with compact support, the two sides act the same way as linear functionals. (Since for any test function $\psi \in C^\infty_c(\mathbb{R})$

$$ \langle(x*u_1),\psi\rangle = -\langle x,(\psi*u_1)\rangle = - \langle x,\psi'\rangle = \langle x',\psi\rangle $$

you see that the two sides are equally defined as distributions, using that a convolution of a test function against a distribution is always well defined and also a smooth function.)

So the equation you wrote down is meaningful in the sense of distributions. And hence you can take $dx/dt$ as a continuous representative (in the equivalent class of distributions) of $x*u_1$. Which I think is about as rigorous as you need it.

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thanks so much for the detailed answer! – AgCl Nov 9 '10 at 1:59

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