# Dirac function

1. Mar 14, 2010

### Aroma2010

Dirac function :(

Hello everyone...

I have some triple with my PDEs course especially with the Dirac function.
How can I prove it is discontinuous function?
I do not know where can I start...

2. Mar 15, 2010

### gato_

Re: Dirac function :(

Well, for a start, the Dirac "function" is not even considered a function. It doesn't qualify for it. There are two approaches two define it. You can considere it as a limit of several continuous sequences of functions, for example:

$$\delta_{n}=\frac{1}{n\sqrt{\pi}} e^{-x^{2}/n^{2}} , n\rightarrow 0$$

is one such sequence. The limit, however, is not defined as a function. The right name is distribution. See http://en.wikipedia.org/wiki/Distribution_(mathematics)

3. Mar 15, 2010

### Redbelly98

Staff Emeritus
Re: Dirac function :(

(Moderator's note: thread moved from "Differential Equations")

Start with the definition of a continuous function. At least one of the conditions is not satisfied, at at least one value of x, for a function to be discontinuous.

4. Mar 15, 2010

### IttyBittyBit

Re: Dirac function :(

It's pretty simple really. Just prove that the left and right limits at 0 are 0 (pretty easy), so in order to be continuous, f(0) must be 0, but it isn't, QED.