Is it possible to define distributions acting on non-smooth functions?

The reason I'm asking is because of the rendering equation $$L_o(x,\omega_o) = \int_\Omega f_r(x,\omega_o,\omega_i) L_i(x,\omega_i) \text{d}\omega_i.$$

In this equation, $f_r$ is the Bidirectional Reflection Distribution Function (BRDF). However, a common BRDF is the Dirac delta distribution (for optically perfect mirrors). This makes me want to define $f_r$ as a distribution. The rendering equation would then look like $$ L_o(x,\omega_o) = \langle f_r(x,\omega_o),L_i(x) \rangle$$ which I think seems nice.

The problem is that since $L_i$ is the incoming light at $x$ from direction $\omega_i$, it is often a non-smooth function. According to the little I know about distributions, they only act on test functions in $C^\infty$ (and with compact support, which isn't a problem). Does this mean that I cannot define the BRDF as a distribution?

What would then be a mathematically rigorous way of stating the rendering equation while allowing $f_r$ to be the Dirac delta distribution?


I think I found what I was looking for: Mollifiers. I should be able to convolve $L_i$ with a Mollifier: $L_{i,\epsilon} = \phi_\epsilon \ast L_i$. This gives me a smooth function and I can then define the rendering equation as $$ L_o(x,\omega_o) = \lim_{\epsilon \to 0} ~\langle f_r(x,\omega_o),L_{i,\epsilon}(x) \rangle$$ Does this look correct?

  • $\begingroup$ Some distributions can act on all sorts of things; in particular, the Dirac delta distribution acts on all functions, period (since all functions can be evaluated at a point). But, for example, the $n^{th}$ derivative of the Dirac delta distribution can only act on $n$-times-differentiable functions. $\endgroup$ – Qiaochu Yuan Aug 10 '15 at 17:07
  • $\begingroup$ Thanks a lot for the reply @qiaochu-yuan! $\endgroup$ – vidarn Aug 10 '15 at 19:21
  • $\begingroup$ However, I am still a bit confused. All the literature I've read have defined distributions as acting on smooth test functions. Admittedly, I've never really understood the purpose of having them smooth, but how come we can just ignore that requirement? $\endgroup$ – vidarn Aug 10 '15 at 19:23
  • $\begingroup$ Having them act on smooth functions presumably makes the theory nicer in various ways, e.g. it means the derivative of a distribution is another distribution, and probably it has nice implications for the Fourier transform. It all depends on what you want to do. $\endgroup$ – Qiaochu Yuan Aug 11 '15 at 7:43
  • $\begingroup$ @QiaochuYuan that makes sense. Thanks a lot for your input! I guess it all depends on which other operations I want to do on the distribution. $\endgroup$ – vidarn Aug 11 '15 at 11:00

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