# A Fun with Deltas and $L^2$: Physicists vs Mathematicians

1. Apr 9, 2016

### stevendaryl

Staff Emeritus

Why do you say that? The function that takes a function $f(x)$ and returns $f(0)$ is a well-defined distribution. Maybe what you mean is that if we identify two functions $f \sim g$ if $\int |f(x) - g(x)|^2 dx = 0$, then the delta-function distribution is ill-defined on the equivalence class. But it's perfectly well-defined on square-integrable functions, if we don't take equivalence classes.

Last edited by a moderator: Apr 13, 2016
2. Apr 9, 2016

### stevendaryl

Staff Emeritus
$\delta(x)$ is not really a function, so $\delta(0)$ is not defined. The only thing that is defined about the delta function is integrals involving it:

$\int_{a}^b \delta(x) f(x) = f(0)$ if the interval contains $x=0$, and is zero, otherwise.

Statements involving $\delta(x)$ that don't explicitly involve integrals can only be made sense of by understanding them in terms of integration.

3. Apr 9, 2016

### rubi

It's not continuous, since $\lVert f - g\rVert = 0$, but $\left| F(f)-F(g)\right| \neq 0$. So it would be an element of the algebraic dual, but not the topological dual space.

4. Apr 9, 2016

### stevendaryl

Staff Emeritus
It's a little confusing. When people talk about $L^2$, they do mean functions modulo equivalence. But it seems to me that we can certainly talk about the set of square-integrable functions without taking equivalence relations.

5. Apr 9, 2016

### stevendaryl

Staff Emeritus
Viewed as functions (as opposed to equivalence classes of functions), $\int |f(x) - g(x)|^2 dx = 0$ does not imply that $f=g$, so I don't see what's the problem with having $F(f) \neq F(g)$.

6. Apr 9, 2016

### rubi

The space of square-integrable functions is usually denoted by $\mathcal L^2$ rather than $L^2$, but it is not a Hilbert space, so it's not suitable for QM.

Your $F$ is well-defined on $\mathcal L^2$. It's just not continuous and distributions are usually defined to be continuous. It's okay to use discontinuous distributions, but you must be aware that many things might not work anymore as expected.

7. Apr 9, 2016

### micromass

That's the thing isn't it. If $f = g$ a.e., why would we expect the integrals $\int f(x)\delta(x)dx$ and $\int g(x)\delta(x)dx$ to be different.
Furthermore, what will be the derivative of $\delta$ in your formalism?
And why did you not put any continuity requirements on the definition of your distribution?

I mean: ok, sure, your definition is consistent, but I don't see it as very useful.

8. Apr 9, 2016

### stevendaryl

Staff Emeritus
That's the whole point--$\int f(x) \delta(x) dx$ is not actually an integral. It's just notation. That notation is defined to be $f(0)$.

I don't understand the question. $\delta(x)$ is not a function. It doesn't have a derivative. It only makes sense in the expression $\int \delta(x) f(x) dx$.

A distribution is not in general an integral. Certain distributions can be defined via an integral, but the delta function is one that can't be.

There are no continuity requirements: $F(f) \equiv f(0)$. That's well-defined for any function that has a value at $x=0$

9. Apr 9, 2016

### micromass

You do realize that no math book adopts the definition you put down in this thread, right?

10. Apr 9, 2016

### stevendaryl

Staff Emeritus
Well, the context of this discussion is the delta function, and there is no way to make sense of a delta function in terms of continuous distributions.

11. Apr 9, 2016

### micromass

Why would you say that? Of course there is...

12. Apr 9, 2016

### stevendaryl

Staff Emeritus
13. Apr 9, 2016

### micromass

You really don't see the difference between what you wrote and what's in the wiki link? For one, the space of test functions in the wiki link is not $\mathcal{H}$, and second they require continuity.

14. Apr 9, 2016

### stevendaryl

Staff Emeritus
Could you elaborate? Because I don't know what you are talking about.

I'm just giving the same definition as Wikipedia, which is this:

As a distribution, the Dirac delta is a linear functional on the space of test functions and is defined by

for every test function φ.

15. Apr 9, 2016

### rubi

The delta distrubution is continuous on the usual test function spaces like $C^\infty_c$ or $\mathcal S$. You can embedd them into $L^2$ by sending each $f$ to its equivalence class. Of course, delta isn't continuous in the Hilbert space norm, but it's continuous in the topologies of the usual test function spaces.

(Edit: Since micromass is active, I'm going to let him continue the argument.)

16. Apr 9, 2016

### micromass

OK, and what is their space of test functions? Also, read the following paragraph which deals about continuity.

17. Apr 9, 2016

### stevendaryl

Staff Emeritus
Look, the functional $F$ defined by $F(f) = f(0)$ is defined on the set of all functions from $R$ to $C$, regardless of whether they are continuous, or square-integrable, or whatever. You might be right that this function is only considered a "distribution" when it is restricted to a suitable collection of functions.

18. Apr 9, 2016

### Staff: Mentor

The limits of integration are any range that includes the point at which δ(x) ≠ 0.

19. Apr 10, 2016

### strangerep

Heh,... gotta love these physicist-vs-mathematician encounters... [SCNR]

(To be fair, the "usual topology on the space of test functions" is far from trivial, imho.)

20. Apr 10, 2016

### Samy_A

As a student, I could get really angry when someone dared to mention the delta function.