1. ## The delta distribution

Hi all,

I am trying to show that the delta distribution cannot be written in terms of an integral. To do this, we assume otherwise and deduce a contradiction.

If it were so, we would have $\phi(0) = \int_{-\infty}^{\infty} \delta(x)\phi(x) dx$ for all test functions $\phi$, so in particular for the test function defined by $e^{-a^2/(a^2 - x^2)}$ if $|x| \leq a$, and zero otherwise, and so we would have

$1/e = \int_{-a}^a \delta(x)e^{-a^2/(a^2 - x^2)} dx$.

The book I am reading then argues that if \delta were a locally integrable function, Lebesgue's general theorem of convergence would tell us that the integral converges to 0.

I cannot see how this can be deduced by the theorem. Can anyone help me out?

2. Consider the sequence of step functions

$f_n(x)=n \cdot \bigg{\chi}_{\left[-\frac{1}{2n},\frac{1}{2n} \right]}$

to approximate the delta distribution. Where chi is the indicator function.

note that

$\lim_{n \to \infty}\int_{\mathbb{R}}f_n=1 \ne \int_{\mathbb{R}}\lim_{n \to \infty}f_n=0$

If the delta distribution was locally integrable we could pass the limit inside.

3. Thanks for your reply. This is useful, but I'm not sure if I explained myself very well. I wish to get a contradiction with the Lebesgue's general theorem of convergence - it is this theorem that I am not sure how to apply, since I do not see how to dominate delta with an integrable function.

4. You're assuming that $\delta$ comes from a locally integrable function, in particular | \delta 1_{[-a,a]} | \leq | \delta 1_{[-1,1]} | for all $a<1$, and the function on the left is integrable.

Another way to do this is by checking that the distributional derivative of the Heaviside function does not come from a locally integrable function.

Edit: Apparently my Latex doesn't work for some reason, so I'll leave it like that.

5. Don't I need to show that delta(x)e^{(-a^2)/(a^2 - x^2)} is dominated by some integrable function g?

I do not see why delta(x)1_[-1, 1] would be integrable - isn't this function just delta(x) again?

6. Originally Posted by measureman
Don't I need to show that delta(x)e^{(-a^2)/(a^2 - x^2)} is dominated by some integrable function g?

I do not see why delta(x)1_[-1, 1] would be integrable - isn't this function just delta(x) again?
delta is integrable (at least locally) by assumption, you're trying to derive a contradiction from this. The exponential is less than one and delta(x)1_[-1, 1] is integrable, so you have your bound.

7. Ah, this makes more sense. I was getting confused between what I was assuming. I'm sure you can tell that I'm fairly new to the DCT.

My question is why do we need the indicator functions? Can't we just use |delta(x)| instead of |delta(x)1_[-1, 1]|?

I also need to show that delta(x)e^{(-a^2)/(a^2 - x^2)} converges pointwise to 0 as a converges to 0. Now if x is not 0, this expression is 0, so of course it is true. The problem I have is with x = 0, how do I justify it here?

Thanks.

8. Originally Posted by measureman
Ah, this makes more sense. I was getting confused between what I was assuming. I'm sure you can tell that I'm fairly new to the DCT.

My question is why do we need the indicator functions? Can't we just use |delta(x)| instead of |delta(x)1_[-1, 1]|?

I also need to show that delta(x)e^{(-a^2)/(a^2 - x^2)} converges pointwise to 0 as a converges to 0. Now if x is not 0, this expression is 0, so of course it is true. The problem I have is with x = 0, how do I justify it here?

Thanks.
The indicator functions are needed because the function representing delta is only assumed locally integrable ie. integrable on every compact set.

For the other question, the dominated convergence convergence theorem only requires convergence almost everywhere. Notice that your exponential doesn't converge to zero, but multiplied by 1_{[-a,a]} it does converge to zero.

9. Originally Posted by Jose27
The indicator functions are needed because the function representing delta is only assumed locally integrable ie. integrable on every compact set.

For the other question, the dominated convergence convergence theorem only requires convergence almost everywhere. Notice that your exponential doesn't converge to zero, but multiplied by 1_{[-a,a]} it does converge to zero.
Thanks. Is the following a valid argument?

Suppose delta is a locally integrable function. It follows from $|\delta(x)e^{-a^2/(a^2 - x^2)}1_{[-a, a]}| <= |\delta(x)1_{[-a, a]}|$ that $\delta(x)e^{-a^2/(a^2 - x^2)}$ is dominated by an integrable function (by assumption). For each a > 0, we have that $\delta(x)e^{-a^2/(a^2 - x^2)}$ is a real-valued measurable function and we can see that $\delta(x)e^{-a^2/(a^2 - x^2)}$ tends to 0 as a tends to 0 almost everywhere. LDCT would imply that $1/e = \lim_{a \to 0} \int_{-a}^a \delta(x)e^{-a^2/(a^2 - x^2)} dx = \int_{-a}^a \lim_{a \to 0} \delta(x)e^{-a^2/(a^2 - x^2)} = \int_{-a}^a 0 dx = 0$.

we can see that $\delta(x)e^{-a^2/(a^2 - x^2)}$ tends to 0 as a tends to 0 almost everywhere
This is not true, that function tends to $\delta(x)$ a.e. The one that goes to 0 a.e. is $\delta (x)e^{-a^2/(a^2 - x^2)}1_{[-a,a]}$. Another thing, your interchange of limits is not valid as written since the integration limits involve 'a'; to fix this notice that for any function f we have $\int_{-a}^a f = \int_{\mathbb{R}}f1_{[-a,a]}$.