# A parabola delta function

1. Apr 27, 2005

### saltydog

Can someone help me prove the following:

$$L=\mathop \lim\limits_{k\to \infty}\int_{-\frac{3}{4k}}^{\frac{3}{4k}} f(x)[-\frac{16k^3}{9}x^2+k]dx=f(0)$$

I'm pretty sure at the limit,

$$-\frac{16k^3}{9}x^2+k$$

becomes a delta function.

Essentially, it's that section of a narow parabola above the x-axis growing taller and more narrow with increasing k. The limits are it's roots and the area is always 1 between the roots.

I can prove the case for the simple square-wave pulse version of delta using the intermediate value theorem for definite integrals but I don't know how to approach the above one. I'd like to show more work but I don't have anymore. It's a hard limit to evaluate for me.

2. Apr 27, 2005

### dextercioby

What limit (presumably after "k") are u implying...?0...?

Daniel.

3. Apr 27, 2005

### saltydog

Hello Daniel,

I'm taking the limit of the value of the integral as k goes to infinity. The parabola then becomes infinitely tall and infinitesimally narrow yet the area beneath it between its roots remains at unity. I've checked it with Mathematica and indeed the limit is f(0), whatever function I use. For example, if f(x)=$x^2+9$, then the limit would be f(0)=9.

I plan to spend some time on it and see what I can get.

4. Apr 27, 2005

### dextercioby

Shouldn't the limit go to 0...?I mean the filtering property of delta-Dirac should be

$$\int_{-\infty}^{+\infty} f(x)\delta(x) \ dx =f(0)$$

in the general case,or

$$\int_{-\epsilon}^{+\epsilon} f(x)\delta(x) \ dx =f(0)$$

for an arbitrary finite real 'epsilon'.Your case doesn't correspond...Make a connection to what i know and thoroughly applied in physics (viz,the 2 eqns written above).

Daniel.

5. Apr 27, 2005

### saltydog

Though someone would bring that up. Thus let:

$$S_k(x)=\left\{\begin{array}{cc}-\frac{16k^3}{9}x^2+k,&\mbox{ if }|x|\leq\frac{3}{4k}\\0, &\mbox{ if }|x|>\frac{3}{4k}\end{array}\right$$

Now, I could just as well have used a variable v which represents the distance between the two roots. This distance goes to 0 as k goes to infinity. Thus, using the definition above and the variable v, I can then say:

$$\int_{-\infty}^{\infty}f(x)S(x)dx=f(0)$$

if:

$$S(x)=\mathop \lim\limits_{v\to 0}S_k(x)$$

With S(x) being the new delta function.

And therefore:

$$L=\mathop \lim\limits_{v\to 0}\int_{-\frac{3}{4k}}^{\frac{3}{4k}} f(x)[-\frac{16k^3}{9}x^2+k]dx=f(0)$$

6. Apr 27, 2005

### dextercioby

There's no "v" in your last equation.The limit is useless and the integral cannot be evaluated,if the function 'f' is even on the integration domain.

Daniel.

7. Apr 27, 2005

### saltydog

v is a function of k. The limit is not useless as any function supplied to it and evaluated as v(k) goes to zero is f(0). I believe S(x) meets all the requirements of being a genuine delta function.

As far as even or odd, evaluate the limit for:

$$f(x)=3x^3$$

The result will be 0 and this is an odd function.

Evaluate the limit for:

$$f(x)=x^2+9$$

The result will be 9 and that's even.

8. Apr 27, 2005

### dextercioby

I am right and confident that your last formula from post # 5 doesn't make any sense whatsoever...That's what i said...

Daniel.

9. Apr 27, 2005

### saltydog

Daniel, I wish to go a step further and propose the following claim which I stand by until proven otherwise:

$$\int_{-\infty}^{\infty}f(x)g(x-a)dx=f(a)$$

If and only if:

$$g(x)\in \mathbb{D}$$

where:

$$\mathbb{D}$$ is the set of base functions which at the limit become Delta Functions and further:

$$S(x)\in\mathbb{D}$$

where S(x) is defined above.

Last edited: Apr 27, 2005
10. Apr 27, 2005

### Hippo

I think I'm close, or something...

Proving this:
$$\mathop \lim\limits_{k\to \infty} \int_{-\frac{3}{4k}}^{\frac{3}{4k}} [k-\frac{16k^3}{9}x^2]f(x)dx=f(0)$$

By Mean Value Theorem,

$$\int_{-\frac{3}{4k}}^{\frac{3}{4k}} [k-\frac{16k^3}{9}x^2]f(x)dx = \frac{3}{2k} [k-\frac{16k^3}{9}c^2]f(c)$$

Where

$$c \in (-\frac{3}{4k}, \frac{3}{4k})$$

By simple squeeze theory,

$$\mathop \lim\limits_{k\to \infty} \rightarrow \mathop \lim\limits_{c\to 0}$$

So

$$\mathop \lim\limits_{k\to \infty} [\frac{3}{2} -\frac{8k^2}{3}c^2]f(c) = [\frac{3}{2} -\frac{8}{3}]f(0)$$

This result is, well, close, but not quite it.

Am I missing something obvious or am I going about it the wrong way?

Last edited by a moderator: Apr 27, 2005
11. Apr 27, 2005

### snoble

I have to admit I've just scanned the thread but looking at the original problem
$$L=\mathop \lim\limits_{k\to \infty}\int_{-\frac{3}{4k}}^{\frac{3}{4k}} f(x)[-\frac{16k^3}{9}x^2+k]dx=f(0)$$
Have you considered the integral as a Lebesgue integral. As a quick glance this looks like exactly the sort of problem that the Lebesgue integral can illuminate. Although you have to consider are you dealing with a measurable function, a continuous function, etc...

Daniel I don't believe there is anything wrong with defining a limit on a function of a variable. It makes perfect sense when you go back to the def'n of a limit. If given any $$\epsilon>0$$ there is a $$\delta>0$$ st if $$||v(x)-k||<\delta$$ then $$||f(x)-L|| <\epsilon$$ then
$$\lim\limits_{v(x)\to k}f(x)=L$$
Although he doesn't seem to define v(x) anywhere: he may have something in mind.

Last edited: Apr 27, 2005
12. Apr 27, 2005

### snoble

Thinking about the problem a bit more it occurs to me that this will only work on a narrow set of functions.
Consider $$f(x) = x\cdot\sin(1/x)$$ except at 0 where f(0)=0. Here you have a continuous integrable function but the above limit doesn't seem to exist (just from some experimental data in maple). Maybe if you limit your functions to differentiable functions or analytic functions but then you are getting really constraining.

13. Apr 28, 2005

### saltydog

I can let:

$$v=\frac{3}{4k}$$

Then the limit becomes:

$$\mathop \lim\limits_{v\to 0}\int_{-v}^{v} f(x)[-\frac{3}{4v^3}x^2+\frac{3}{4v}]dx$$

which I claim is equal to f(0).

Doing so and as stated above, using the intermediate value theorem, I can reduce the limit to:

$$\mathop \lim\limits_{v\to 0}[\frac{3}{2}(f(\xi v)(1-\xi^2))$$

which if this limit is to converge to f(0), then somehow, $\xi=\frac{1}{\sqrt{3}}$

That's where I'm at so far.

Also, with regards to restraining the set of functions applicable as indicated by Snoble, are other delta function representations likewise constrained and if so then it is a limitations of the function class and not the particular function used above.

What I find particularly interesting is letting:

$$f(x)=k\sqrt{x}+a$$

In that case, the integral does not have a real solution between the limits of -v and v, yet the limit exists and is equal (I propose), to a.

14. Apr 28, 2005

### saltydog

Hello Snoble,

I'm having a problem with that function.

Seems to me it's not continuous at f(0) and also:

$$\int_{-a}^a \frac{x}{Sin[1/x]}dx$$

does not exist for any values of a. Am I correct?

If so, then perhaps the range of functions applicable to the delta relationship should be constrained to "functions which are continuous" at x=0. I'm not really familiar with delta theory. Perhaps this is a constraint already imposed on the subject.

Edit: I should add that Mathematica returns "indeterminate", the following integral:

$$\int_{-\infty}^{\infty} \frac{x}{Sin[1/x]}\delta(x)dx$$

Where $\delta(x)$ is the Dirac delta function.

Last edited: Apr 28, 2005
15. Apr 28, 2005

### saltydog

I'm making progress showing:

$$\mathop \lim\limits_{v\to 0}\int_{-v}^{v}f(x)[\frac{3}{4v}-\frac{3x^2}{4v^3}]dx=f(0)$$

According to the Intermediate Value Theorem for definite integrals:

$$\int_{-v}^{v}h(x,v)dx=2vh[\xi v,v]$$

for some $-1<\xi<1$

Where I'm using h(x,v) to represent the entire integrand.

If I substitute this into the expression, I'm left with:

$$\int_{-v}^{v}f(x)[\frac{3}{4v}-\frac{3x^2}{4v^3}]dx=\frac{3}{2}f(\xi v)[1-\xi^2]$$

Thus, in order to show the delta relation above, must show:

as $v\to 0, \xi\to\frac{1}{\sqrt{3}}$

Well, I can't yet show this, however, numerically, when I use:

$$f(x)=3x^2+4$$

And evaluate the integral for 0<v<1, and plot the corresponding value of $\xi$ (actually the negative), I obtain the plot below. This plot "appears" to converges to $\frac{1}{\sqrt{3}}$ as indicated by the horizontal line marking such. I realize this is not proof but offers incouragement towards that direction.

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16. Apr 28, 2005

### snoble

sorry I wasn't clear. The $$\cdot$$ means times and not divide. $$f(x) = x\cdot sin(1/x)$$ (except that f(0)=0) is the classic misbehaving continuous everywhere function (including at 0). Mathematica should have a problem with the integral of the dirac function times $$x\cdot sin(1/x)$$ since you have to separately define it for 0. But the integral $$\int_{-a}^a x\cdot sin(1/x)$$ is defined for every $$a$$ even if you don't define it at 0.

Try seeing if mathematica has either a 'piecewise' or a characteristic function.

Also try taking a sequence for $$\int_{-\frac{3}{4k}}^{\frac{3}{4k}} f(x)[-\frac{16k^3}{9}x^2+k]dx$$ with the k's getting larger. Or plot it with say k from 1000 to 1100. You will see why I don't think the limit converges. Of course this isn't a proof it doesn't converge. Just some pretty sugestive evidence.

So what does this counter example mean? It means if you are going to prove something you are going to need the differentiablitity of f at 0.

17. Apr 28, 2005

### saltydog

Hello Snoble,

May I ask you what is:

$$\int_{-\infty}^{\infty} x\cdot sin(1/x)\delta(x)dx$$

If it is 0,then how is that determined other then by arbitrarilly defining the function to be 0 at zero? That is, how can it be determined by taking a limit of a function sequence? I realize Arildno is preparing an exposition on the matter and I patiently await his posts. I also appreciate your kind comments on the matter but I still see no difference between the parabola delta function I'm proposing above and other Dirac sequences proposed for the same job. I mean, if the other ones don't converge with this particular function then I could not expect S(x) above to do so.

18. Apr 28, 2005

### Data

19. Apr 28, 2005

### saltydog

Yes, yes, I'm keeping up and very much appreciate his efforts. I'm patient and can tolerate being wrong. I have questions but don't wish to interfere with the presentation.

20. Apr 28, 2005

### Data

You appear to be essentially right, don't worry

21. Apr 28, 2005

### snoble

To be honest I have not come across the dirac delta that often. I too will be reading adrilno's write up since I'm sure I'll learn something. But here's what I do know
$$\int_{-\infty}^{\infty} x\cdot sin(1/x)\delta(x)dx$$ is not defined since $$x\cdot sin(1/x)$$ is not defined at 0. But if you define $$f(x)=x\cdot sin(1/x)$$ when $$x\ne 0$$ and f(0)=a then $$\int_{-\infty}^{\infty} f(x)\delta(x)dx=a$$. If a=0 then f(x) is continuous everywhere. I was just noting that it does seem to be a counter example to your original statement and you should keep counter examples in mind when trying to understand a concept.

22. Apr 28, 2005

### Hippo

Could someone tell me if I'm making any serious logical mistakes?

Where $$x = \frac{3}{4}s$$

$$\int_{-\frac{3}{4k}}^{\frac{3}{4k}} \left[ k-\frac{16k^3}{9}x^2 \right] f(x) \ dx = \int_{-\frac{1}{k}}^{\frac{1}{k}} \frac{3}{4} \left[ k-{k^3}s^2 \right] f \left( \frac{3}{4}s \right) \ ds$$

By Mean Value Theorem, if the indefinite integral is continuous* on [-1/k, 1/k] and differentiable* on (-1/k, 1/k), then

$$\int_{-\frac{1}{k}}^{\frac{1}{k}} \frac{3}{4} \left[ k-{k^3}s^2 \right] f \left( \frac{3}{4}s \right) \ ds = \frac{3}{2} \left[ 1-{k^2}c^2 \right] f \left( \frac{3}{4}c \right)$$

where

$$c \in \left(-\frac{1}{k}, \frac{1}{k} \right)$$

It suffices to provide a c such that

$$|c|<|\frac{1}{k}|$$

and

$$\mathop \lim\limits_{k\to \infty} \frac{3}{2} \left[ 1-{k^2}c^2 \right] f \left( \frac{3}{4}c \right) = f(0)$$

So, let

$$c=\frac{1}{k \sqrt{3}}$$

Then

$$\frac{3}{2} \left[ 1-{k^2}c^2 \right] f \left( \frac{3}{4}c \right) = f \left( \frac{\sqrt{3}}{4k} \right)$$

and

$$\mathop \lim\limits_{k\to \infty} f \left( \frac{\sqrt{3}}{4k} \right) = f(0)$$

* Differentiability and Continuity

Let

$$I(s) = \int \frac{3}{4} \left[ k-{k^3}s^2 \right] f \left( \frac{3}{4}s \right) \ ds$$

Then

$$\frac {\ dI}{\ ds}(p) = \frac{3}{4} \left[ k-{k^3}p^2 \right] f \left( \frac{3}{4}p \right)$$

which exists $$\forall p \in \left[ -\frac{1}{k}, \frac{1}{k} \right]$$

so long as f(3p/4) does.

Continuity on the closed domain follows from differentiability on it.

Last edited by a moderator: Apr 29, 2005
23. Apr 29, 2005

### saltydog

Thanks a bunch Hippo. I shall spend time going through it meticulously however that will take time for me.

24. Apr 29, 2005

### arildno

Be VERY careful about how to define an integral representation of the Dirac sampling functional in the case of discontinuous f's (that is, $$a\neq{0}$$)!!

Your result will not only depend upon the set of comparison/weight functions, but typically also, you'll need to define a particular sequence of F-functions converging to the discontinuous f as well...

Last edited: Apr 29, 2005
25. Apr 29, 2005

### saltydog

Alright Hippo, I think I have a problem with the proof:

The expression above is not valid for a constant c but rather c will change as a function of k. Thus, I believe the statement:

Is not valid. What do you think?