Is There a Simpler Definition for the Dirac Delta Function?

[Swapnil]

https://www.physicsforums.com/showthread.php?t=73447

I saw the above tutorial by arildno and looked at how he defined the Dirac Delta "function" as a functional. But isn't there a more easier way to do this. I have seen the following definition in a lot of textbooks.

$$\delta(t) \triangleq \lim_{\epsilon \to 0} \frac{1}{\epsilon} \Pi\Big(\frac{t}{\epsilon}\Big)$$

where $$\Pi(t)$$ is the gate function and is defined as
$$\Pi (t) := <br /> \begin{cases}<br /> 0 & \mbox{ for } |x| > \frac{1}{2} \\<br /> \frac{1}{2} & \mbox{ for } |x| = \frac{1}{2} \\<br /> 1 & \mbox{ for } |x| < \frac{1}{2},<br /> \end{cases}$$

What's wrong by defining the delta function in this way?

 

[Hurkyl]

Mainly because that limit of real-valued functions does not exist.


However, physics has this weird notational convention that makes everything an operator -- I've never seen anyone try to explicitly state it, but it's evident in how they manipulate different kinds of expressions. When a physicist writes that expression, what they are really writing is the operator pointwise defined as:

$$\delta(t) [ f(t) ] \triangleq \lim_{\epsilon \rightarrow 0}<br /> \frac{1}{\epsilon} \Pi\left( \frac{t}{\epsilon} \right) [ f(t) ]$$

And since $\Pi$ is representable by an actual function, the latter operator application is an honest-to-goodness integral:

$$\Pi\left( \frac{t}{\epsilon} \right) [ f(t) ] =<br /> \int_{-\infty}^{+\infty} \Pi\left( \frac{t}{\epsilon} \right) f(t) \, dt$$

Of course, physicists like to write the left-hand side as an integral too, even though it really isn't:

$$\int_{-\infty}^{+\infty} \delta(t) f(t) \, dt \triangleq \delta(t) [ f(t) ]$$

(unless you get into measure theory, and define $\delta(t) \, dt$ to be a certain measure. But if we did that, then we wouldn't be defining it via the limit expression above!)

 

[arildno]

Swapnil:
What I tried to do in that tutorial, was to give ONE fairly rigorous way to make sense of the Dirac Delta function.

The aim of the tutorial is just that, and being thus limited, many interesting issues are not at all addressed there. For example, you won't find any evolved, mature distribution theory in it.

 

[cliowa]

And since $\Pi$ is representable by an actual function, the latter operator application is an honest-to-goodness integral:

$$\Pi\left( \frac{t}{\epsilon} \right) [ f(t) ] =<br /> \int_{-\infty}^{+\infty} \Pi\left( \frac{t}{\epsilon} \right) f(t) \, dt$$

Just for the record: That's a definition of the operator, right?

 

[Swapnil]

And since $\Pi$ is representable by an actual function, the latter operator application is an honest-to-goodness integral:
$$\Pi\left( \frac{t}{\epsilon} \right) [ f(t) ] =<br /> \int_{-\infty}^{+\infty} \Pi\left( \frac{t}{\epsilon} \right) f(t) \, dt$$

Where is this coming from? Is it a result of something or is it a definition?

 

[Hurkyl]

Where is this coming from? Is it a result of something or is it a definition?

Hrm, good question. Definition, I suppose.

 

[Swapnil]

When a physicist writes that expression, what they are really writing is the operator pointwise defined as:

$$\delta(t) [ f(t) ] \triangleq \lim_{\epsilon \rightarrow 0}<br /> \frac{1}{\epsilon} \Pi\left( \frac{t}{\epsilon} \right) [ f(t) ]$$

And since $\Pi$ is representable by an actual function, the latter operator application is an honest-to-goodness integral:
$$\Pi\left( \frac{t}{\epsilon} \right) [ f(t) ] =<br /> \int_{-\infty}^{+\infty} \Pi\left( \frac{t}{\epsilon} \right) f(t) \, dt$$

So essentially, the delta function can be treated as an operator with the following definition:
$$\delta(t) [ f(t) ] \triangleq \lim_{\epsilon \rightarrow 0}<br /> \frac{1}{\epsilon} \int_{-\infty}^{+\infty} \Pi\left( \frac{t}{\epsilon} \right) f(t) \, dt$$

right? ( I don't know why you included the definition in two steps.).

 

[Hurkyl]

Right. And because one typically defines these operators to operate on certain sets of continuous functions, you can show that's equivalent to the simpler definition of delta:

$$\delta(t) [ f(t) ] = f(0).$$