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what is the product of delta function with itself ? what is the dot product with itself ?

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In the standard setup, neither of these is defined. See en.wikipedia.org/wiki/Distribution_(mathematics) . –  Qiaochu Yuan Dec 3 '10 at 20:56
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product with itself / square mathoverflow.net/questions/48067/… –  Pratik Deoghare Dec 3 '10 at 21:03
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@crasic: Why doesn't it seem believable? Convolution with $\delta(x-x_1)$ is translation by $x_1$. Now for any function $f(x)$, compare $(f(x)*\delta(x-x_1))*\delta(x-x_2)$ with $f(x)*(\delta(x-x_1)*\delta(x-x_2))$. –  Rahul Dec 3 '10 at 21:23
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@J. M.: To me special functions is something completely different - but I can not put my finger on exactly what special functions is. –  AD. Dec 4 '10 at 7:49
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@Rajesh D: Could you give more context? I have made an edit to explain that you cannot define $\delta^2$ meaningfully. –  Jonas Teuwen Dec 4 '10 at 15:46

3 Answers 3

up vote 16 down vote accepted

A distribution is actually a linear functional on the space of compactly supported infinitely differentiable functions (the so called "test functions"). A function $f$ is compactly supported if $\overline{\{x : f(x) \neq 0\}}$ is compact (the overline denotes the closure).

The $\delta$-distribution is a linear functional such that for all $\phi \in C_c^\infty(\mathbb{R}^n)$ we have that $\langle \delta, \phi \rangle = \phi(0)$.

When you want to compute the product of distributions the problem is that you don't have a property which you would really like to have, that is associativity. So for distributions $\alpha$, $\beta$ and $\gamma$ we usually have that $(\alpha \cdot \beta) \cdot \gamma \neq \alpha \cdot (\beta \cdot \gamma)$. Wikipedia gives an example. However this does not really turn out to be a problem in applications. What we do have is convolution.

When we want to do convolution we prefer a smaller class of distributions (for example because on the smaller class the Fourier transform of a distribution in this class is again a distribution in this class). This actually has a rougher class of test-functions, as test functions here we take the Schwartz functions, that are the smooth functions of which the function itself and all its derivatives are rapidly decreasing. $f$ is said to be rapidly decreasing if there are constants $M_n$ such that $|f(x)| \leq M_N |x|^{-N}$ as $x \to \infty$ for $N = 1,2,3,\ldots$.

To begin defining the convolution we first define what the convolution of Schwartz-function is with a tempered distribution. Let $f$ be our tempered distribution, then we can show that the following definition actually makes sense: $$\langle \phi * f, \psi \rangle := \langle f, \tilde{\phi} * \psi \rangle$$ where $\tilde{\phi}(x) = \phi(-x)$. Note that the RHS is well-defined. Convolution is a nice thing, we can see that if we start with a tempered distribution and convolute it with a test function, the result will be smooth. Now, $L_1 * L_2$ is the unique distribution $L$ with the property that $L * \phi = L_1 * (L_2 * \phi)$. We can show that this is commutative.

Fine, now note that $\delta * \phi(x) = \phi(x - y)|_{y = 0} = \phi(x)$. So we see that $\delta * \delta = \delta$.

If you want me to comment on the dot product of distributions, you first would have to explain what you mean with that.

So far for this short digression on distributions.

EDIT: Okay, you want to compute $\delta^2$. Let $\phi_n$ be an approximation to the identity and let it converge to $\delta$ in the sense of distributions, but $\phi_n^2$ does not converge at all since the integral against a test function that does not vanish at the origin blows up as $n \to \infty$.

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Here it is stated that: $\int_{-\infty}^\infty \delta (\xi-x) \delta(x-\eta) \, dx = \delta(\xi-\eta).$ Does the integral make the difference? Or the arguments of the two $\delta$s? –  draks ... May 7 '12 at 21:54
    
@draks It is some kind of tensor product right? –  Jonas Teuwen May 7 '12 at 22:13
    
What makes you think that? To me it looks like a "standard" multiplication... –  draks ... May 8 '12 at 6:47
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You don't multiply them in the same point. –  Jonas Teuwen May 8 '12 at 8:20
    
OK, can I square $\sqrt{\delta(x)}$? Or better $\lim_{a \to 0} \sqrt{\delta_a}$, e.g. $\sqrt{\delta_a(x)}=\exp(-x^2/\color{red}{2}a^2)$ –  draks ... May 8 '12 at 8:35

I think the answer provided by Jonas is correct, but Jonas assumed "product" meant convolution as this is normally the only binary operation that usually make sense with the Dirac delta function.

So if by product you mean convolution then: δ∗δ=δ

If by "product" you meant point-wise multiplication, then the answer is: Undefined.

The usual approach is to treat δ as the limit of some nascent delta function. See Delta Function for examples of these nascent delta functions. So if you multiply two nascent δ functions and then take the limit, the result will vary depending on the particular pair of nascent δ functions selected. In most cases the integral of the point-wise product of two nascent δ functions is zero as the limit is taken, in some cases the integral is 1, in other cases the value of the integral tends to infinity as the limit of nascent δ is taken.

For example If you multiply the rectangle nascent δ with the nascent δ that is a triangle pulse, then integral of the resulting cubic tends to zero as the interval tends to 0. If you multiply the sinc version of the nascent δ with itself, then the integral is equal to the frequency of the sinc function. In this case the integral tends to infinity as the frequency of the sinc function tends to infinity.

Because of this, the original answer given (The question makes no sense) is the best because the term "product" in its common meaning (e.g. point-wise multiplication of two functions) leads to contradictory results for the limit process. Using the term "product" when you mean convolution (the only operation that produces meaningful results) is at best confusing.

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Here is a heuristic to suggest that it will be difficult to define the square of the delta function.

The Fourier transform has the property that it takes the convolution of two functions to the product of their Fourier transforms, and vice versa, ie, it takes the product of two functions to their convolution.

Remember that the Fourier transform of the delta function is the constant function($=1$). Now suppose that $\delta^2$ exists. Then its Fourier transform would be the convolution of two constant functions. Such a convolution would shoot to infinity at every point. Even the theory of distributions can't handle such kind of stuff. So how would you imagine the inverse Fourier transform of this? How would it make sense?

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