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I'm studying for a qualifying exam and in our study group someone asked the question whether the delta function is in $L^2$ spaces.

My argument is that it is; since the delta function function can be approximated by a sequence of $L^2$ functions (say, Gaussian curves with decreasing spread), and $L^2$ spaces are complete, the delta function must be included in them as well, even though the delta function is not a proper function.

This is still a question of controversy in our group, though. Is my thinking correct?

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    $\begingroup$ I say no.${}{}$ $\endgroup$
    – Angina Seng
    Commented Aug 1, 2017 at 18:16
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    $\begingroup$ You would need to prove that your approximating sequence has a limit in $L^2$, i.e. that it's Cauchy. Is it? $\endgroup$
    – Mr. Chip
    Commented Aug 1, 2017 at 18:16
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    $\begingroup$ Dirac delta can be approximated by $L^2$ function in weak-* sense, but this does not tell whether this approximation is also $L^2$ sense. One way of seeing that $\delta$ is not in $L^2$ is to actually show that $$\sup\{ |\langle \delta, \varphi \rangle| : \varphi \in C_c^{\infty}(\mathbb{R}), \ \|\varphi\|_2=1\} = \infty.$$ $\endgroup$
    – Sangchul Lee
    Commented Aug 1, 2017 at 18:25
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    $\begingroup$ Another way to approach the problem is to prove that no $L^2$ function $f$ satisfies $\varphi(0) = \int f(x) \varphi(x) dx$ for all test functions $\varphi$. It is easy to build a sequence $\varphi_n$ for which the integral tends to 0 while $\varphi_n(0)$ does not. $\endgroup$
    – Gribouillis
    Commented Aug 1, 2017 at 19:13
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    $\begingroup$ The Dirac delta is not in any $L^p$ space since it isn't a function. But it is close to be in $L^1$, since "$\int |\delta(x)| \, dx = 1 < \infty$". $\endgroup$
    – md2perpe
    Commented Aug 1, 2017 at 19:40

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This reasoning is not correct. If you want to use completeness, you would need that your sequence is Cauchy in $L^2$ (which it isn't). Furthermore, as you already mentioned yourself, the delta function is not a function (it is a distribution) and is therefore not in $L^2$.

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    $\begingroup$ @DominiqueR.F. The distributions that can be said to be in $L^2$ are given by functions. $\endgroup$
    – md2perpe
    Commented Aug 1, 2017 at 19:35

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