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Let $a,b$ be two real-numbers such that $a \neq b$. Let $\iota(x \leq a)$ be the Heaviside step function in variable $x$ and $\delta_b(x)$ be the Dirac Delta function centered around $b$.

I have two questions:

  1. Is the following correct? $$\int \iota(x\leq a)\delta_b(x) \ dx = \iota(b\leq x).$$ From what I have read regarding distributions, defining the multiplication of distributions is tricky and there is no straightforward way in which the above integral might be defined. Intuitively, however, I don't see why the above doesn't work? Since $a\neq b$ is given, the Heaviside function is essentially a constant around $b$, and even though Heaviside functions aren't test functions, the Delta function $\delta_b$ should be able to operate on $\iota$.

  2. Is the following correct? $$\int\delta_{f(x)}(y)\delta_a(x) \ dx = \delta_{f(a)}(y),$$ where $f$ is a smooth function. Here again I am multiplying two distributions which I believe are not well defined. However, intuitively the above equation seems to make sense.

Could someone please help me out with this. Are the above equations not defined? If so, why? If they are, is there a way to think about them rigorously without having to go beyond the standard theory of distributions.

sp1122
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    Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Apr 06 '24 at 21:49
  • The first question was asked several times already... see e.g. https://math.stackexchange.com/questions/4771005/evaluating-int-p-fp-deltapa-thetap-f-a-theta-a-for-heaviside/4771049#4771049. In general, test functions only need to be continuous near the point where the Dirac delta acts. Mathematically, this is the Riesz representation theorem: the Dirac delta at a point $a$ defined as a distribution (so a linear form on $C^\infty_c$ test functions) can be extended in a unique way to a bounded linear form on functions continuous in any fixed neighborhood of $a$. – LL 3.14 Apr 06 '24 at 22:13

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