I remember the square of the delta function was investigated by Schwinger in QED. There were some strange generalized functions investigated by Shina Tan in his famous paper "Large momentum part of a strongly correlated Fermi gas".  

As far as the square-root of delta, I'm sure it is orthogonal but I'm skeptical it's L2 norm is 1.  It might still be useful but in a larger space of limits of functions than the L2 class.  

Why are you skeptical about the L2 norm? Shouldn't it simply be the (sqrt of the) integral of a delta? Am I missing something?

Consider the representation of the delta function as the rectangle of height H and width w.  Let w->0 while forcing wH=1.  If I take the square root of H this gives an area of w H^1/2 =1/H^1/2->0.  

The reason for the limit I chose is because you presumably want to do some integration and get a result such as \int f(x) \delta(x-y) dy =f(y).  For your square root delta function I assume you would expect \int f(x) (\sqrt\delta(x-y))^2 dy =f(y).  These functions only have meaning in these integral formulas.  If you use your limit above you just get the regular delta function.  

@Remi, are you taking the square root of the function value and its domain at the same time??

@Clifford But as for the L2 norm you have a square inside the integral, i.e. the square of the square root of the the delta, i.e. the Delta itself. That's why the L2 norm should give 1. I agree the L1 norm is zero as you prove with the rectangle representation.

I think Remi has the correct idea, you could define the delta as the limit of a function with a well defined square root. To get the delta out of the limit correctly than you would need to define that the function integrates to 1, in which case the L^2 norm of the square root would be 1 by construction. In this way you could at least construct a function whose square approaches the Dirac delta, and whose L2 norm is 1.

   Let me contribute to the discussion with the attached article. If you take the square root of f_n(x), you get (I think) f_k (x)/sqrt (modulus x) for all  integer x.  

I am a bit sceptical on how showing the area (L1 norm) is zero implies the L2 norm is zero. A priori it doesn't seem like there should be simple relationship between the two.  In fact there are a lot of functions where the L1 norm is zero but the L2 norm is finite, for example x*exp(-x^2). Am I missing something here?

I did forget the absolute value in the definition, sorry bad example.

Take a rectangular function of height H and width 1/H. This is, roughly, a delta. Then the square root, which here is unproblematic, has height H^(1/2) and width still 1/H. So the L1 is H^(1/2), which tends to zero. The L2 norm of the square root is the L1 norm of the delta function, hence the square root is normalised. At least in the approximation, it belongs to L2. Of course, the approximation by a rectangular function is not orthogonal to all L2 functions, but in the limit it becomes this: indeed the functions to which the rectangular function is not orthogonal are those with support inside the width, which is of order H. So ``in the limit'' the square root is an element of L2 with norm 1 but orthogonal to all L2, that is, zero.  This contradiction indicates that taking the limit has taken you out of L2. On the other hand, as an element of L1, since its norm is zero, the square root is zero.

Of course, this does not mean that the square root can be replaced by zero everywhere: in particular, the square of the square root is the delta function, which is not the same as the square of zero. But whenever you use the square root in an expression where an L1 function would make sense, then you can replace the square root by zero. An expression such as int f(x)^2 dx is not an expression that makes sense for an L1 function, since such integrals generally diverge. On the other hand, int f(x)g(x) dx does, if g(x) is bounded, so whenever you use the square root of  a delta in the place of f(x), it can be replaced by 0.

Finally, let me concur with Robert: Colombeau algebras will go a long way towards clarifying your doubts. The idea is to extend the space of distributions, and to have several different concepts of identity. This relates to what I said above: viewed as an element of L1, the square root is zero, but it is not so in general.

   A small correction to my previous message. If you define f_n (x) as in rq.(1) of the attached paper, then sqrt [f_n (x)] is equal to x. exp(-k) for n even, and to (modulus x).exp (- k) times (modulus x). exp(- 1/2) for n odd, for any natural number k. The idea is that delta is the limit of f_n (x) when n goes to infty; then, sqrt (delta(x)) is the limit of sqrt (f_n(x)) when n tends to infty.

  This is just a suggestion.

   Good luck.  

we need a distribution whose squarring is Dirac delta....

OK, guys; I guess you are right.

Objects like sqrt (delta(x)) are not defined in Schwartz-Sobolev's theory of distributions, because a Schwartz distribution (that is, a generalized function) is defined as a LINEAR functional on the space of test functions. See, for example, R.P. Kanwal, Generalized Functions:Theory and Techniques,  Academic Press, 1983, page 25.  However, in real life properties of some solutions of some differential equations can be IDENTIFIED as those of an object like sqrt(delta(x)). As always, new branches of math emerge if there is a call from physics/engineering. Indeed, in recent decade mathematicians introduced non-linear generalized functions (or non-linear distributions). Currently, they develop a rigorous theory of non-linear distributions. A good introduction and references can be found in J.F. Colombeau, Non-linear generalized functions: their origin, some developments and recent advances, Sao Paulo Journal of Mathematics, V. 7. No. 2, pp.201-239 (2013). Good luck!

I think I misread part of the question.  The L2 norm should be one since the square is the delta which has the L1 norm (for positive definite representations) equal to 1.  This means the L1 norm of the square root of delta (for such reps) does vanish so it is not interesting in the usual use of generalized functions.  It would be interesting to see what the application is.  In QED the square of delta arises in some scattering problems and we make sense of that by the normalization so the divergence of the result is not a problem.  If you want to see some clever use of generalized functions see the Tan paper.  

Remi,

I forgot to mention that in the reference I gave above, J.F. Colombeau, Non-linear generalized functions: their origin, some developments and recent advances, Sao Paulo Journal of Mathematics, V. 7. No. 2, pp.201-239 (2013), there is a discussion concerning the definition and properties of sqrt(delta(x)). It begins on page 204. While the paper looks a bit difficult to read for a non-math scientist or student, try it and you will see that it is fun. You will understand most of discussion concerning your question. All in all, properties of sqrt(delta(x)) are different from those of delta(x).

Minas Balyan: not a very illuminating answer.

This is OK if f(x) = 1 for all x.  

I thought that integral (from minus infty to plus infty) was equal to f(0)...  

Hi

See eqs. (33) to (38) of Preprint The Limit-in/out Approach to the Dirac Delta (with an Altern...