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I am reading Alexander Altland and Jan von Delft's Mathematics for Physicists. They introduce the delta function as follows. $\delta_{y}(x)$ is the function such that for all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ $$\int_{\mathbb{R}}\delta_{y}(x)f(x)\,dx=f(y)$$Any constant function $f(x)=c$ immediately gives me $$\int_{\mathbb{R}}\delta_{y}(x)\,dx=1$$ The authors then claim that the delta function vanishes everywhere except at $x=y$, otherwise we can choose an $f$ to give us a contradiction. Quote:

If $\delta_{y}(x)$ were non-vanishing for $x\neq{}y$, it would be possible to devise functions $f(x)$ such that the integral $\int_{\mathbb{R}}\delta_{y}(x)f(x)\,dx$ yields a value different from $f(y)$. Think about this point. (272)

Assume that $x_{0}\neq{}y$ and $\delta_{y}(x_{0})\neq{}0$. How would I define $f$ to get the promised contradiction?

maibaita
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    You can't really do that; if it were finite but nonzero at finitely many points, then you would not get any contradiction. You could argue "well but what if it's continuous?" but that's a pretty flimsy objection considering that the delta function is already not continuous. So I would not take this description too seriously. $\delta_y$ takes a function which is continuous at $y$ and returns $f(y)$, and a common syntax for writing this "application" is $\int_{-\infty}^\infty f(x) \delta_y(x) dx$. That's the only thing you can really take seriously up to this point. – Ian Feb 18 '21 at 23:45
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    It is not a function and there is no way to really associate it to a well defined function. There are limiting arguments that say that it is an infinite spike but that again isn't well defined as a function. This seems like a fool's errand to me. – Cameron Williams Feb 18 '21 at 23:47
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    Thirding the points already made. Best to think of it just as something that operates on functions in the specified way in integration. You can't "rigorize" a treatment that is not already rigorous. Who knows what they were thinking. – leslie townes Feb 18 '21 at 23:48
  • Consider $$f(x) = \begin{cases}1 & x\neq x_0\ 0 & x = x_0 \end{cases}$$ then $$\int_{\mathbb{R}}\delta_{x_0}(x)f(x),dx=f(x_0) = 0$$ this might help for the intuitive argument. – RyanK Feb 18 '21 at 23:48
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    @RyanK I would actually say that this $f$ is not in the domain of $\delta_{x_0}$, but your description has some merit at the level of approximate identities (all approximate identities will see your $f$ as the zero function). – Ian Feb 18 '21 at 23:49
  • Ahh thanks for pointing this out, I was not aware of that! – RyanK Feb 18 '21 at 23:50

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