Dirac Delta as the limit of a Gaussian

[bdforbes]

Show that

$$\stackrel{lim}{\alpha \rightarrow \infty} \int^{\infty}_{-\infty}g(x)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx = g(0)$$

where g(x) is continuous.

To use the continuity of g(x) I started from

$$\left|g(x)-g(0)\right|<\epsilon$$

and tried to put it in into the integral:

$$\left| \int^{\delta}_{-\delta}g(x)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx - \int^{\delta}_{-\delta}g(0)\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx \right| \leq \int^{\delta}_{-\delta}\left|g(x)-g(0)\right|\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx < \epsilon \int^{\delta}_{-\delta}\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} dx$$

But I'm not sure where this gets me.

 

[Avodyne]

This is not helpful because x is not always close to 0 as you integrate. Can you see a change variables that would help?

 

[Dick]

This is not helpful because x is not always close to 0 as you integrate. Can you see a change variables that would help?

I don't see how a change of variable would help. The OP now has to argue that the contributions to the integral in the limit from outside of [-delta,delta] can also be bounded by epsilon by taking alpha sufficiently large. Exactly how you would technically do this depends on what additional conditions you want to put on g(x). And you do need something. If g(x)=exp(x^4) then the limit doesn't even exist.

 

[bdforbes]

I'm getting close to the proof, starting off as I did above. I think it is safe to assume the integral should exist for any value of alpha, because we are using this as a formal definition of the Dirac Delta, so the integral is an improper Riemann integral. So there are conditions on g(x). I just need to find a nice bound for

$$\int^{\infty}_{\delta}\sqrt{\frac{\alpha}{\pi}}e^{- \alpha x^2}dx$$

 

[bdforbes]

I've run into a bit of a problem. I was forgetting that g(x) could go to infinity as x goes to +- infinity, as long as g(x)e^(-ax^2) goes to zero quickly enough. Having a finite value of

$$\stackrel{sup}{x\in\Re}\left|g(x)-g(0)\right|$$

is essential to my proof, because it allowed me to focus my attention on

$$\int^{\infty}_{\delta}\sqrt{\frac{\alpha}{\pi}}e^{- \alpha x^2}dx$$

which I managed to find a bound for. However now I must also consider

$$\int^{\infty}_{\delta}\left|g(x)-g(0)\right|\sqrt{\frac{\alpha}{\pi}}e^{- \alpha x^2}dx$$

It's intuitive that we can make this integral as small as we want by increasing alpha, but I can't figure out how to prove this.

EDIT:

I guess we can just consider

$$\int^{\infty}_{\delta}\left|g(x)\right|\sqrt{\frac{\alpha}{\pi}}e^{- \alpha x^2}dx$$

 

[bdforbes]

$$\psi(x) = \sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2}$$

$$\stackrel{lim}{\alpha \rightarrow \infty} \psi(x) = 0$$ for x != 0

So for a given interval $$(\delta, \infty)$$, we can reduce the values of $$\left|g(x)\right|\psi(x)$$ as much as we want by increasing alpha enough.
Can this fact be used to prove that the integral goes to zero?

EDIT:

Would the above method require g(x) to have absolute continuity? That might be a deal breaker.

 

[Dick]

No, you don't need absolute continuity or anything. Let's write psi(a,x)=sqrt(a/pi)*exp(-ax^2). Now suppose the integral of psi(a0,x)*g(x) converges on [delta,infinity) for some a0 (you'll need to assume something like this to get anywhere at all). Can you find a way to pick an a1>a0 such that |psi(a1,x)*g(x)|<|psi(a0,x)*g(x)|/2 for all x in [delta,infinity)?? That would do it, right?

 

[bdforbes]

Yes, I'm sure we could pick such an a_1, or prove that such a choice is possible. However, isn't this the same as showing that |g(x)|psi(a,x) goes to zero everywhere in (delta, infinity) as a goes to infinity? How can we say anything about the limit of the integral?

 

[Dick]

Showing a function converges to zero everywhere doesn't show that it's integral converges to zero. Take f(x)=exp(-(x-a)^2) as a->infinity. That converges to zero everywhere pointwise, but the integral doesn't converge to zero. You just have to show that your integral isn't one of this group. If you can pick that a_1, then you've shown that. And I guess you are right, this is a form of uniform convergence.

 

[bdforbes]

Thanks, I assumed there was a theorem like that which would help. How could we prove this theorem?

 

[bdforbes]

Okay I solved it. I started by proving I could find a_1>a_0 such that |g(x)|psi(x,a_1)<(1/2)|g(x)|psi(x,a_0) as you suggested, with a_1 a function of a_0 and delta only. Then I looked at the integrals over a finite interval (delta,R), and proved that the proper integral for a_1 is half the proper integral for a_0, then taking R to infinity we find that the same relation applies for the improper integral. This gives us a sequence {a_n} that allows us to reduce the value of the integral arbitrarily provided n is large enough, thus by definition the limit is zero.

At all steps I reduced things to limits of finite sums, so I'm confident that I have satisfied any convergence conditions. I'm going to try and write it up concisely now.

Thanks for your help Dick.

 

[Dick]

Right. That's what I was thinking of. But you don't need finite interval number R. Having a function going to zero and an integral going to zero on a finite interval STILL doesn't guarantee convergence of the improper integral. Look at the exp(-(x-a)^2) example. I would just assume that g(x)psi(x,a_0) converges. The theorem isn't true for an arbitrary continuous function g, the integrals may not exist. I think it would be fair to assume they do exist before you deal with the limit.

 

[bdforbes]

I agree that we should assume that, since the question kind of doesn't make any sense if the integral doesn't converge. I wasn't using the finite interval (delta,R) to prove convergence, just as an intermediate step to make sure I was doing everything properly.

 

[Dick]

Ok, just checking. I think you've got this one. Good job.