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A common way to characterize the dirac delta function $\delta$ is by the following two properties:

$$1)\ \delta(x) = 0\ \ \text{for}\ \ x \neq 0$$

$$2)\ \int_{-\infty}^{\infty}\delta(x)\ dx = 1$$

I have seen a proof of the sifting property for the delta function from these two properties as follows:

Starting with

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

for some "sufficiently smooth" function $f$, since $\delta(x - t) = 0$ for $x \neq t$ we can restrict the integral to some epsilon interval around $t$

$$\int_{-\infty}^{\infty}\delta(x-t)f(x)\ dx = \int_{t-\epsilon}^{t+\epsilon}\delta(x-t)f(x)\ dx$$

On this infinitesimal interval, $f$ is "approximately constant" and so we can remove it from the integral

$$\int_{t-\epsilon}^{t+\epsilon}\delta(x-t)f(x)\ dx = f(t)\int_{t-\epsilon}^{t+\epsilon}\delta(x-t)\ dx = f(t)$$

This proof seems a little too hand wavy for me. The points I find problematic are in quotations. What is meant by "sufficiently smooth" in this case? Is continuous enough? Also, how exactly is the extraction of the function from the integral done rigorously, without just assuming that it is "approximately constant"? I have seen this proof done with non-standard analysis and I understand that the delta function is by nature a rather "hand wavy" object so that a rigorous proof using these two properties may not even exist. Still I ask if anyone can perhaps make the above proof rigorous or offer a new proof without appealing to non-standard analysis.

(I'm not too sure what tags to include for this question. If anyone could retag for me that'd be much appreciated

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First a small remark : the dirac delta is not strictly speaking a function, it's called a distribution. It's often defined as being the distribution such that $\int f(x) \delta(x) dx = f(0)$. Using that definition, your equality follows from a change the variable in the integral (from x to x−t). This definition also gives you the properties you state. On the other hand, if you start from your characterization, then I think the proof you give can easily be made rigorous if $f$ is continuous at $t$ (which gives a precise quantification for "approximately constant"). –  Joel Cohen Oct 16 '11 at 11:18
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Well, as you mention, no truely rigorous treatment can be given with such a description of the Delta Dirac function - no such function actually satisfies those requirements. Thus, I won't take too much effort to make the below too precise.

Where it says "sufficiently smooth", it doesn't actually need anything there at all! Whatever $f$ is, as long as it is finite almost everywhere, the product with that delta function will be $0$ away from a neighbourhood of $t$, so you can restrict the integral like that.

For the extraction of $f$, being continuous in a closed neighbourood of $x=t$ is enough. If $f$ is continuous through $ [ t-\epsilon, t+ \epsilon ] $ then by the Extreme value theorem it attains a maximum and minimum in that interval, call them $ M $ and $m$ respectively.

Then since $ m \leq f(x) \leq M$ is that range, $$ m=\int^{t+\epsilon}_{t-\epsilon} m \delta(x-t) dx \leq \int^{t+\epsilon}_{t-\epsilon} f(x) \delta(x-t) dx \leq \int^{t+\epsilon}_{t-\epsilon} M \delta(x-t) dx = M.$$

Now as $\epsilon \to 0$, both $m$ and $M$ go to $f(t)$ as $f$ is continuous so by the Squeeze theorem, $$ \int^{t+\epsilon}_{t-\epsilon} f(x) \delta(x-t) dx \to f(t) $$ as $\epsilon \to 0$.

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Theory of Distributions by J. Ian Richards and Heekyung Youn gives rigorous proofs of things like this without need for functional analysis, topology, or measure theory.

Definitely the proof you give is extremely "hand-wavy". In particular, the two defining properties you give cannot be taken literally if one conceives of a function as something where you put in a number $x$ and get out a number $f(x)$. To take those two properties is intuitive but very very "hand-wavy".

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Yea I thought so. I have seen this made more or less rigorous via non-standard analysis, although with the added condition that this delta "function" be non-negative. I was just curious whether a characterization of the function like this can be done rigorously or not. Thank you for the book recommendation. –  EuYu Oct 17 '11 at 1:58
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