# Multiplication of two impulse function $\delta(t)\cdot \delta(t)=?$

If a impulse function is mulplied with a function $f(x)$ then the formula will be apply $$f(x)\delta(x-a)=f(a)\delta(x-a)$$ so from this formula mulplication of two impulse function will be

$$\delta(t)\delta(t)=\delta(0)\delta(t)$$

But,what if i take two impulse function as the approximation of standard signal (rectangular or gaussian) and mulplied them directy then i will get again an impulse function of same property.

What is wrong with this two?

• Strictly speaking the function multiplying the generalized function $\delta(t)$ should be a sufficiently smooth continuous one. Otherwise the multiplication is undefined. You can have a look at the generalized function theory and distribution theory books. Commented Oct 1, 2017 at 17:33
• See the mathoverflow question Is square of Delta function defined somewhere? Commented Oct 2, 2017 at 6:58

The property

$$f(t)\delta(t-t_0)=f(t_0)\delta(t-t_0)\tag{1}$$

is only valid for a function $f(t)$ that is continuous at $t=t_0$. Since the Dirac delta impulse $\delta(t)$ is not a function (it is a distribution) and since $\delta(t)$ is not continuous, property $(1)$ does not hold (and does not make sense) for $f(t)=\delta(t)$.

The product $\delta(t)\cdot \delta(t)$ is undefined. Also note that the quantity $\delta(0)$ is meaningless. Since $\delta(t)$ is not a function you cannot determine its value for any value of its argument.

• have a look at the original question and consistent usage of $\delta(0)$ in its solution. (where $\delta(0)$ is used in both time domain and Fourier domain computation of an improper (divergent) convolution) Commented Oct 2, 2017 at 9:12
• @Fat32: OK, but I still don't think that $\delta(0)$ makes any sense. $\delta(0)=\infty$ may be an intuitive way to think about it, but it is not correct. Commented Oct 2, 2017 at 16:05

You can use the following (weak) argumentation to deduce that the result is interpreted as the (not very meaningful statement of) $$\delta(t) \cdot \delta(t) = \delta(0) \cdot \delta(t)$$

Define the impulse as a limit of the following pulse function

$$\delta(t) = \lim_{\Delta \to 0} \delta_{\Delta}(t)$$ where $\delta_{\Delta}(t)$ is a rectangular function with the definition that

$$\delta_{\Delta}(t) = \begin{cases} 0 &,\ t < 0 \\ \frac{1}{\Delta} &,\ 0 \le t < \Delta \\ 0 &,\ \Delta \le t \end{cases}$$

Then you can define the multiplication as: \begin{align} \delta(t)\delta(t) &= \left( \lim_{\Delta \to 0}\delta_{\Delta}(t) \right) \left( \lim_{\sigma \to 0}\delta_{\sigma}(t) \right)\\ &= \lim_{\Delta \to 0} \lim_{\sigma \to 0} \left[ \delta_{\Delta}(t) \delta_{\sigma}(t) \right]\\ &= \lim_{\Delta \to 0} \left( \delta_{\Delta}(t) \lim_{\sigma \to 0} \delta_{\sigma}(t) \right)\\ \text{now observe that}\\ \delta_{\Delta}(t) \lim_{\sigma \to 0} \delta_{\sigma}(t) &\approx \delta_{\Delta}(0) \lim_{\sigma \to 0} \delta_{\sigma}(t) = \delta_{\Delta}(0) \delta(t) \end{align} where the equality holds in the limit. Then proceed with

\begin{align} \delta(t)\delta(t) &= \lim_{\Delta \to 0} \left( \delta_{\Delta}(0) \delta(t) \right) \\ \delta(t)\delta(t) &= \left( \lim_{\Delta \to 0} \delta_{\Delta}(0) \right) \delta(t) \\ \delta(t)\delta(t) &= \left( \delta(t)|_{t=0} \right) \delta(t) \\ \delta(t)\delta(t) &= \delta(0) \delta(t) \\ \end{align}

• So why is it "not very meaningful"? All the math looks perfectly OK to me. Commented Oct 1, 2017 at 20:42
• @DilipSarwate, does "$\delta(0)$" look perfectly OK? Commented Oct 2, 2017 at 4:52
• and i would like to see anyone successfully convolve these two functions: $$x(t) = 1 \qquad \forall t \in \mathbb{R}$$ and $$h(t) = 1 \qquad \forall t \in \mathbb{R}$$ Commented Oct 2, 2017 at 4:59
• @DilipSarwate not meaningful in the classical (19th century) mathematical sense... Classical point set topology is fundamentally based on smoothness and density. So that the concept of limit, integral and derivate makes sense. It cannot be used to model a blackhole nor a quark however... More modern mathematics deals with distributions in a more consistent way but I'm not sure of its justifications, which however are no less meaningful than the justifications of smoothness! Commented Oct 2, 2017 at 9:23
• @robertbristow-johnson The point is that the limit does not exist at $t=0$ (the value of $\delta_{\Delta}(t)$ at $t=0$ is $\frac{1}{\Delta}$ which diverges to $\infty$ as $\Delta \to 0$) and even if the limit exists, we cannot claim that the "value" of $\delta(t)$ at $t=0$ equals this limit without assuming that $\delta(t)$ is continuous at $t=0$. In short, Fat32's "argument" perpetuates the myth that $\delta(t)$ is an ordinary function. That he knows better is shown in his comment above as well as in his comment on the OP's question. Commented Oct 2, 2017 at 15:15

Well, since

$$\pi^2\delta(0)^2=2\int_0^\infty t dt-\frac1{12}$$ we have $$\delta(0)^2=\frac2{\pi^2}\int_0^\infty t dt-\frac1{12\pi^2}$$

So, we can write $$\delta(x)^2$$ as

$$\delta(x)^2=\frac1{\pi^2}\int_{-\infty}^\infty |t| \cos(xt) dt-\frac{0^{|x|}}{12\pi^2}$$

• This is complete nonsense Commented Dec 1, 2020 at 10:04
• @Jazzmaniac what exactly? This does not work for you? Commented Dec 1, 2020 at 10:08
• Do you actually believe that what you wrote here is proper math? If you do, then there's no point in argueing. Commented Dec 1, 2020 at 10:10
• No, that's not an explanation. It's just more incoherent, hand-waving and inconsistent impressionistic "math". This is very far from a rigorous argument. My comment above still stands. Commented Dec 1, 2020 at 10:55
• @Jazzmaniac if you see some inconsistencies, you are welcome to point them. There are definitely which I know, but the do not touch this result. Commented Dec 1, 2020 at 11:16