# Square root of the delta function

• I
Is square root of delta function a delta function again?

$$\int_{-\infty}^\infty f(x) \sqrt{\delta(x-a)} dx$$

How is this integral evaluated?

PeroK
Homework Helper
Gold Member
You could try approximating the delta function by a sequence of finite spike functions and see what happens.

This is actually related to Bell's theorem. It is not the solution to the OP : searching A s.t. $$\int A (a,\lambda)A (\lambda,b)d\lambda=-cos (a-b)$$

If we make the assumption $$A (a,\lambda)=f (a-\lambda)$$ and let $$u=a-\lambda$$ we get a convolution :

$$\int f (u)f (a-b-u)du=-cos (a-b)$$

Using the Fourier convolution theorem :

$$2F^2 (\omega)=-\delta (\omega+1)-\delta (\omega-1)$$

Hence the searched function f is given by

$$f (x)=\frac {i}{\sqrt {2}}\int e^{ikx}\sqrt {\delta (k+1)+\delta (k-1)}$$

Bell's theorem shows that there is no such f.

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Addendum : no such f in the range -1 to 1

∫∞−∞f(x)√δ(x−a)dx​
\int_{-\infty}^\infty f(x) \sqrt{\delta(x-a)} dx

How is this integral evaluated?
not every sequence of mathematical symbols makes sense

You could aslo think it in the functional sense : ##\int\delta(x-y)f(y)dy=f(x)=:\Delta_x(f)##

Then the root of delta would be a semi evaluation of f , which you could compute by semi integrating, using Cauchy's formula for repeated integration extended to real values of fractional calculus :

##\sqrt{\Delta_x}(f)=\frac{1}{\Gamma(1/2)}\lim_{y\rightarrow\infty}\int^y\delta(x-y)f(t)(y-t)^{-1/2}dt##

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• WWGD
Error : my last formula is wrong since it contains a function depending on the limits of integration hence Cauchy's formula for repeated integral cannot be so easily used

anuttarasammyak
Gold Member
I have never met such a formula before. If it can exist, I suppose
$$\int f(x)\sqrt{\delta(x-a)}dx=\int \frac{f(x)}{\sqrt{\delta(x-a)}}\delta(x-a)dx=\frac{f(a)}{\sqrt{\delta(0)}}=0$$

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• Delta2
wrobel
You could try approximating the delta function by a sequence of finite spike functions and see what happens.
and then check that what it is happened does not depend on a sequence

jasonRF
Gold Member
EDIT: I just realized that this thread is three years old...

To me it seems like ##\sqrt{\delta(t)}## probably doesn't exist. Here is my thinking (this is not a proof).

First, as suggested by PeroK, start with one particular sequence of finite spikes that converges to a delta function. Let ##p_n(t) = n## for ##-\frac{1}{2n}\leq t\leq\frac{1}{2n}##, and zero everywhere else. If ##\sqrt{\delta(t)}## exists, then I would hope that the sequence ##\sqrt{p_n(t)}## would converge to it. Let ##\phi## be a continuous test function, then
$$\begin{eqnarray*} \lim_{n\rightarrow\infty}\int \phi(t) \sqrt{p_n(t)} & = & \lim_{n\rightarrow\infty} \sqrt{n}\int_{-1/2n}^{1/2n} \phi(t) dt \\ \lim_{n\rightarrow\infty} \sqrt{n}\,\phi(\tau)\int_{-1/2n}^{1/2n} dt \end{eqnarray*}$$
for some ##\tau \in [-\frac{1}{2n}, \frac{1}{2n}]##. Then obviously we find that the limit is zero. So we have ##\sqrt{p_n(t)}\rightarrow 0##, even though ##p_n(t)\rightarrow \delta(t)##. This suggests that either ##\sqrt{\delta(t)}=0## or it doesn't exist.

Now I want to look at the product of ##\sqrt{\delta(t)}## with itself. We usually don't talk about multiplying distributions since in general the product of two distributions is not a distribution. However, multiplication of two distributions is allowed in those cases where the product is a distribution. In this case, if ##\sqrt{\delta(t)}## is a distribution then obviously it can be multiplied by itself to give the delta distribution. Also, if ##\sqrt{\delta(t)}## is a distribution then it has a Fourier transform that we will denote by ##\mathcal{F}\left[\sqrt{\delta(t)}\right]=F(\omega)##. Now we can use the convolution theorem for Fourier transforms $$\begin{eqnarray*} 1 & = & \mathcal{F}\left[{ \delta(t)}\right] \ & = & \mathcal{F}\left[\sqrt{ \delta(t)}\sqrt{ \delta(t)}\right] \ & = & \frac{1}{2\pi}F(\omega)\ast F(\omega). \end{eqnarray*}$$
I don't know whether a distribution ##F(\omega)## exists that is a constant when convolved with itself. If ##F(\omega)## doesn't exist, then neither does ##\sqrt{\delta(t)}##; if it does exist then clearly ##F(\omega)\neq 0##. However, this seems to contradict our earlier finding that suggests ##\sqrt{\delta(t)}=0## which would imply ##F(\omega)=0##.

So I suspect that ##\sqrt{\delta(t)}## doesn't exist. Or if it does exist, it doesn't mean what the notation suggests it should mean.

jason

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• wrobel
$$\delta(x)$$ is not a function but a distribution according to Distribution Theory, I do not think of any possible case of finding it under a squareroot?

wrobel
About multiplication of generalized functions see Colombeau theory of generalized functions

• jasonRF
jasonRF
Gold Member
About multiplication of generalized functions see Colombeau theory of generalized functions
Do you know of any references on Colombeau theory that are appropriate for non-mathematicians? It looks intriguing, but the references that popped up with a quick google search seem to require a little more mathematical knowledge/sophistication than I have. There are many "simple" treatments of the Schwartz theory of distributions - such as the book by Strichartz (where I first learned it). Is there anything equivalent for the Colombeau theory, or does it inherently require more tools to properly explain?

thanks,

Jason

anuttarasammyak
Gold Member
Supplement to post #9

As eigenfunction of x, ##\delta(x)## have the relation
$$f(x) \delta(x) = f(0) \delta(x)$$

If we allow f be not only normal function but also distribution ##\delta^{a}(x)##
$$\delta^{a}(x)\delta(x)=\delta^{a}(0)\delta(x)$$

Let a=-1/2
$$\delta^{1/2}(x)=\delta^{-1/2}(0)\delta(x)$$
so it is delta function times the coefficient ##\delta^{-1/2}(0)=0##. So
$$\delta^{1/2}(x)=0$$ in the similar sense with
$$x\delta(x)=0$$
when multiplied by normal function f of f(0) ##\neq \infty## and integrated.

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wrobel