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

Staff Emeritus

## Main Question or Discussion Point

@stevendaryl Why would you take your test space $\mathcal{H}$ as the square integrable functions? That doesn't work at all. The Dirac delta function would be ill-defined. The distributions you get by taking $\mathcal{H} = L^2$ are just the elements of $L^2$ again.
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.

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Staff Emeritus
Thanks Steven, I did read the wiki definitions before I asked the question. I think where i am having difficulty is at x = 0. By definition, f(x) is infinity at zero. I see that as a discontinuity. There are no limits of integration. Where is the Δx? The limit of a function view implies there is a Δx→ 0. (Or the W parameter in the example). I might be being pig headed about this.
$\delta(x)$ 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.

rubi
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.
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.

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.
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.

Staff Emeritus
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.
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 $F(f) \neq F(g)$.

rubi
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.
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.

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 $F(f) \neq F(g)$.
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.

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.
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.

Staff Emeritus
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.
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 $f(0)$.

Furthermore, what will be the derivative of $\delta$ in your formalism?
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.

And why did you not put any continuity requirements on the definition of your distribution?
There are no continuity requirements: $F(f) \equiv f(0)$. That's well-defined for any function that has a value at $x=0$

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 $x=0$
You do realize that no math book adopts the definition you put down in this thread, right?

Staff Emeritus
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.
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.

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

Staff Emeritus
The definition in terms of distributions is completely standard:
https://en.wikipedia.org/wiki/Dirac_delta_function#As_a_distribution
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.

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

rubi
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.
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.)

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

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

jtbell
Mentor
There are no limits of integration.
The limits of integration are any range that includes the point at which δ(x) ≠ 0.

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.)

Samy_A
Homework Helper
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.)

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

strangerep

As a student, I could get really angry when someone dared to mention the delta function.
Well? It's a generalized function, isn't it? Don't mathematicians like to generalize?

[Edit: @Kevin McHugh: Sorry if this little side-exchange is a distraction from your main topic. Let us know if your original question is still inadequately answered.]

As a student, I could get really angry when someone dared to mention the delta function.
The delta function is a function though, just not one $\mathbb{R}\rightarrow \mathbb{R}$

vanhees71
Gold Member
2019 Award
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.
The Dirac $\delta$ distribution cannot be a dual vector in $L^2$, because these are one-to-one mapped to the square integrable functions themselves, because the dual of the separable Hilbert space is equivalent to the Hilbert space itself. The $\delta$ distribution is defined on a dense subspace (like the space of rapidly decreasing $C^{\infty}$ functions). It's dual is larger than the Hilbert space. In context of quantum theory it's a functional defined on the domain of the position operator as an essentially self-adjoint operator.

bhobba
Mentor
Staff Emeritus
The Dirac $\delta$ distribution cannot be a dual vector in $L^2$, because these are one-to-one mapped to the square integrable functions themselves, because the dual of the separable Hilbert space is equivalent to the Hilbert space itself.
But isn't it the case that in the $L^2$ Hilbert space, two functions $f$ and $g$ are considered equal if $\int |f(x) - g(x)|^2 dx = 0$? So the Hilbert space is actually a set of equivalence classes of functions, rather than a set of functions. So for a distribution $F$ to be well defined on that Hilbert space, it must be that $F(f) = F(g)$ whenever $\int |f(x) - g(x)|^2 dx = 0$.