Is the delta function absolutely integrable?
vv
really! Even the man on the street knows that int delta(t) dt = 1.
The question I have is, what about int |delta(t)| dt, i.e., the
integral of the absolute value of the delta function ? There are
sequences that tend to delta that are absolutely integrable, but what
got me confused is that the sinc function (sin pi x / pi x) also tends
to the delta function as it grows skinnier and taller; BUT, this guy
is not absolutely integrable. So, what's the verdict on absolute
integrability of the Dirac delta? Thanks!
--VV
Chip Eastham
In the distributional sense, yes, the (unit) delta
function is absolutely integrable. It is not,
however, square integrable.
regards, chip
A N Niel
<1e980977-1f0b-4ad0...@y10g2000prf.googlegroups.com>, vv
<vana...@netzero.net> wrote:
You have to define what you mean by this.
Theorems with "absolutely integrable" as hypothesis are for functions,
not for distributions, so even if you get an answer to your question,
it will not be relevant for those theorems.
W^3
<c9759386-f146-40d5...@y19g2000yqy.googlegroups.com>,
Chip Eastham <hard...@gmail.com> wrote:
What does it mean for a distribution to be absolutely integrable, or
square integrable?
Chip Eastham
> In article
> <c9759386-f146-40d5-8ce5-0ab71e1c8...@y19g2000yqy.googlegroups.com>,
Hi, Wade:
Well, it's a great question and one the OP ducked,
but I'll try not to take too trivial of a route
and give an answer that has some math behind it.
Generally speaking we define norms on distributions
by "duality" since distributions are mere bounded
(linear) functionals on spaces of actual functions,
which we will take to be C_0^oo(R), i.e. infinitely
continuously differentiable real functions with
compact support on the real line.
Letting F be a linear functional on this space, we
can define a "dual norm" ||F||_q with respect to
an L^p norm on "test functions" g as follows:
||F||_q = sup F(g)
where the supremum is taken over all g in C_0^oo(R)
whose L^p norm is (less than or) equal to 1.
Let's consider the L^2 norm and square integrability
first. The Riesz Representation Theorem tells us
that any linear functional bounded with respect to
the L^2 norm of test functions can actually be
represented by an L^2 function (in a conventional,
measurabilty sense; not your cheesy distributional
imitations). Since the delta function cannot be so
represented, I'm taking refuge in that fact to claim
it's not square integrable (i.e. the L^2 norm is
+oo, as I defined it above).
The L^1 norm is then my criteria for "absolutely
integrable", and here we invoke the dual of the
regular function space L^1(R) to get the L^oo norm
on test functions. Then for delta function F we
have ||F||_1 = 1.
regards, chip
David C. Ullrich
What's below may be what _you_ mean by the
phrase. But it's not a correct answer to Wade's
question "what does it mean?". In standard
terminology there's simply no such thing as
an "absolutely integrable" distribution.
No. If F is in L^1 then the L^1 norm is indeed
given by duality this way. It does not follow that
the L^1 norm of the delta function is what you
say - there's simply no such thing as the L^1
norm of the delta function.
Yes, you can use language any way you like, as long
as you're careful to make clear when you mean something
non-standard by a term. But it's very important to be
clear about that - applying standard theorems about
absolutely integrable functions to things that you juat
happen to _call_ absolutely integrable functions is
going to lead to contradictions.
(For example: It's a theorem that if f is absolutely
integrable and epsilon > 0 there exists delta > 0
such that for any set E with mu(E) < delta we have
int_E |f| d mu < epsilon...)
>
>regards, chip
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
David C. Ullrich
wrote:
>Which ever way you slice, this delta function is monkey business,
>really! Even the man on the street knows that int delta(t) dt = 1.
The man on the street knows a lot of things. If we're careful about
our definitions it's not true that the integral of the delta function
is 1. Because the "delta function" is not a function, it's a
distrubution, and distributions simply don't have integrals.
>The question I have is, what about int |delta(t)| dt, i.e., the
>integral of the absolute value of the delta function ?
This makes less sense - a distribution does not have an absolute
value.
I don't know what level you're at mathematically. In various
books on, for example, differential equations, the author
talks about the delta function as though it were a function
with an integral. But it's not so - they just talk that way
in those books because they assume the reader doesn't
have the mathematical background to follow the real
explanation.
>There are
>sequences that tend to delta that are absolutely integrable, but what
>got me confused is that the sinc function (sin pi x / pi x) also tends
>to the delta function as it grows skinnier and taller; BUT, this guy
>is not absolutely integrable. So, what's the verdict on absolute
>integrability of the Dirac delta? Thanks!
>
>--VV
David C. Ullrich
Chip Eastham
> On Mon, 27 Jul 2009 12:59:10 -0700 (PDT), Chip Eastham
> >Generally speaking we define norms on distributions
Hi, David:
Don't sugarcoat it! Is my definition of L^1 norms
of distributions ill-founded, useless, or merely
nonstandard and thus a threat to unwary newsgroup
readers?
I think I was clear enough in saying the "delta
function" is not actually a function but a
distribution (aka "generalized function").
The first question to ask about a generalization
is whether it is sound (well-defined), and then
whether it is useful. As for being nonstandard,
we are already on dangerous ground with the
"delta function" nomenclature, but it has
survived, perhaps an abuse of notation, because
(I'm guessing) of its utility as a pedantic
device (insert deprecating comment about
engineers and physicists here). The most we
can hope for is to gently insert "generalized
function" into the conversation as a step to
winging them over to thinking of distributions.
There are numerous norms defined on spaces of
distributions which generalize familiar norms
on spaces of functions. Unsurprisingly these
definitions are in terms of dual norms.
To that extent my approach is standard, and I
don't think I've given an ill-posed definition
of the L^1 norm of a distribution. As for the
L^2 norm, I'm sure you can think of more
applications of the Riesz Representation Thm.
than I can.
E.g. the Fourier transform of a (tempered)
distribution can be defined, and we want to
be assured that if we have a distribution
that can be represented by a function, the
two Fourier transforms "agree".
> Yes, you can use language any way you like, as long
> as you're careful to make clear when you mean something
> non-standard by a term. But it's very important to be
> clear about that - applying standard theorems about
> absolutely integrable functions to things that you juat
> happen to _call_ absolutely integrable functions is
> going to lead to contradictions.
>
> (For example: It's a theorem that if f is absolutely
> integrable and epsilon > 0 there exists delta > 0
> such that for any set E with mu(E) < delta we have
> int_E |f| d mu < epsilon...)
It's a nice example that perfectly illustrates
the need for extra care when introducing a
generalization. There's a tendancy to assume
"everything goes through", but sometimes it
ain't so. But it doesn't stop mathematicians
from generalizing; it seems to whet their
appetite for open questions about what the
"best" generalization is, or what altered
sense the original theorem best carries over
to the generalized context.
Here I'd suppose the best we can do is treat
the delta function (or other L^1 distribution)
as giving rise to an associated finite positive
measure mu. In the specific case of the delta
function this could be the unit weight discrete
measure at the origin.
regards, chip
David C. Ullrich
It's _incorrect_ as an answer to the original
question. Because the question was not about
whether or how we could extend the standard
terminology, the question was whether the
delta function _is_ absolutely integrable.
>I think I was clear enough in saying the "delta
>function" is not actually a function but a
>distribution (aka "generalized function").
>
>The first question to ask about a generalization
>is whether it is sound (well-defined), and then
>whether it is useful. As for being nonstandard,
>we are already on dangerous ground with the
>"delta function" nomenclature, but it has
>survived, perhaps an abuse of notation, because
>(I'm guessing) of its utility as a pedantic
>device (insert deprecating comment about
>engineers and physicists here). The most we
>can hope for is to gently insert "generalized
>function" into the conversation as a step to
>winging them over to thinking of distributions.
>
>There are numerous norms defined on spaces of
>distributions which generalize familiar norms
>on spaces of functions. Unsurprisingly these
>definitions are in terms of dual norms.
And some of those norms _do_ have the property
that if a distribution has finite norm then it is
absolutely integrable. The norm you've defined
does not have this property. Even leaving aside
the question of misleading the OP into thinking
the answer to his question is something other than
"no", it's awesomely hard to see _why_ we
should want to extend the definition of "absolutely
integrable" to include distributions for which
that norm is finite.
I mean, if someone asked whether pi was an
even integer one could explain that with a
suitable redefinition of "even integer" the
answer would be yes. For example we could
declare that any real number was an even
integer. What purpose would be served by
that? Assuming your answer to that is "none":
what's the difference between that and the
present case?
(Especially since there is _already_ a
standard name for the distributions with
finite Chip norm: they are (the distributions
defined by) complex (Borel) measures.)
David C. Ullrich
Chip Eastham
> It's _incorrect_ as an answer to the original
> question. Because the question was not about
> whether or how we could extend the standard
> terminology, the question was whether the
> delta function _is_ absolutely integrable.
I stand corrected. But your answer (no) seems
to rest simply on the fact that the delta function
is not a function, something I highlighted while
trying to dig out a slightly deeper "fact".
> I mean, if someone asked whether pi was an
> even integer one could explain that with a
> suitable redefinition of "even integer" the
> answer would be yes. For example we could
> declare that any real number was an even
> integer. What purpose would be served by
> that? Assuming your answer to that is "none":
> what's the difference between that and the
> present case?
pi is no kind of integer. A better analogy
would be if someone asks whether 1 + sqrt(5)
is an even integer. In that case we could
well say it is not a rational integer, but
it is an algebraic integer and indeed a
multiple of 2 times an algebraic integer.
The concept of even integer can be extended
in this well-defined/nontrivial way.
> (Especially since there is _already_ a
> standard name for the distributions with
> finite Chip norm: they are (the distributions
> defined by) complex (Borel) measures.)
Thanks, I do like to learn standard names for
things.
regards, chip
G. A. Edgar
Ullrich <dull...@sprynet.com> wrote:
> I mean, if someone asked whether pi was an
> even integer one could explain that with a
> suitable redefinition of "even integer" the
> answer would be yes.
Odd, actually. I vaguely recall some textbook where "even" was defined
in the usual way, then "odd" is defined as any number that is not even.
Somehow this had been placed in a section on the real number system, so
that ... pi is odd!
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
vv
>
> I don't know what level you're at mathematically. In various
> books on, for example, differential equations, the author
> talks about the delta function as though it were a function
> with an integral. But it's not so - they just talk that way
> in those books because they assume the reader doesn't
> have the mathematical background to follow the real
> explanation.
I am not a math major but merely use delta functions in the few times
they show up. Calling the delta as a "function" has probably caused
me to raise the incorrect question ("Is the delta function absolutely
integrable?"). After reading the replies I realize that the matter
requires a familiarity and understanding at a level that I don't
possess.
--VV
David C. Ullrich
<ed...@math.ohio-state.edu.invalid> wrote:
>In article <9upu65dloldvb1tka...@4ax.com>, David C.
>Ullrich <dull...@sprynet.com> wrote:
>
>> I mean, if someone asked whether pi was an
>> even integer one could explain that with a
>> suitable redefinition of "even integer" the
>> answer would be yes.
>
>Odd, actually. I vaguely recall some textbook where "even" was defined
>in the usual way, then "odd" is defined as any number that is not even.
>Somehow this had been placed in a section on the real number system, so
>that ... pi is odd!
That _is_, um, curious (pun narrowly avoided).
David C. Ullrich
If you weren't aware that distributions with
finite Chip norm are simply complex measures
then you probably didn't see why calling them
L^1 distributions is really a bad thing to do.
The point is that there's this whole theory
(see a book on measure theory) on the
_connection_ between complex measures
and L^1 functions. A complex measure is
an L^1 function if and only if ...., any
complex measure has a unique decomposition
as an L^1 function plus a singular measure,
etc. If you take that theory and redefine L^1
to include all complex measures the effect
really is analogous to taking number theory
but deciding that all numbers are even.