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<Moderator's note; thread forked from https://www.physicsforums.com/threads/question-regarding-the-dirac-delta-function.866220/>

Why do you say that? The function that takes a function [itex]f(x)[/itex] and returns [itex]f(0)[/itex] is a well-defined distribution. Maybe what you mean is that if we identify two functions [itex]f \sim g[/itex] if [itex]\int |f(x) - g(x)|^2 dx = 0[/itex], then the delta-function distribution is ill-defined on the equivalence class. But it's perfectly well-defined on square-integrable functions, if we don't take equivalence classes.

micromass said:@stevendaryl Why would you take your test space ##\mathcal{H}## as the square integrable functions? That doesn't work at all. The Dirac delta function would be ill-defined. The distributions you get by taking ##\mathcal{H} = L^2## are just the elements of ##L^2## again.

Why do you say that? The function that takes a function [itex]f(x)[/itex] and returns [itex]f(0)[/itex] is a well-defined distribution. Maybe what you mean is that if we identify two functions [itex]f \sim g[/itex] if [itex]\int |f(x) - g(x)|^2 dx = 0[/itex], then the delta-function distribution is ill-defined on the equivalence class. But it's perfectly well-defined on square-integrable functions, if we don't take equivalence classes.

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