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David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250

Posted: Sat Jul 15, 2006 4:20 pm    Post subject: Re: The Fundamental Theorem of Calculus

On Fri, 14 Jul 2006 14:46:46 -0700, The World Wide Wade

 Quote: In article , David C. Ullrich wrote: On Thu, 13 Jul 2006 09:48:58 EDT, Maury Barbato mauriziobarbato@aruba.it> wrote: David C. Ullrich wrote: [...] If F' = f is in L^1[a,b], then the Lebesgue integral of f over [a,b] equals F(b)-F(a). True. But it's worth pointing out that one needs to be very careful here: This is true if F is differentiable at _every_ point and the derivative is integrable. Seems like it should be pointed out, since "F' in L^1" is often taken to mean something very different from that. Out of curiosity: what is this alternative meaning? A few comments on that. First, elements of L^1 are not actually functions, they're equivalence classes of functions where two functions are identified if they agree almost everywhere. True if you're in the metric space L^1, but not true for statements like "If f is in L^1(R), then lim_h->0+ 1/h * int_(x,x+h) f = f(x) for a.e. x."

Huh?

************************

David C. Ullrich
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Sat Jul 15, 2006 6:42 pm    Post subject: Re: The Fundamental Theorem of Calculus

 Quote: A few comments on that. First, elements of L^1 are not actually functions, they're equivalence classes of functions where two functions are identified if they agree almost everywhere. True if you're in the metric space L^1, but not true for statements like "If f is in L^1(R), then lim_h->0+ 1/h * int_(x,x+h) f = f(x) for a.e. x." Huh?

If f is in L^1 the metric space, then the sentence in quotes is
nonsense; the integrals of f are well defined, but the values of
f are not. My point is that the notation L^1 is used in two ways:
in the sense you mentioned and as as a space of actual functions.
David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250

Posted: Sun Jul 16, 2006 12:19 pm    Post subject: Re: The Fundamental Theorem of Calculus

On Sat, 15 Jul 2006 11:42:13 -0700, The World Wide Wade

 Quote: A few comments on that. First, elements of L^1 are not actually functions, they're equivalence classes of functions where two functions are identified if they agree almost everywhere. True if you're in the metric space L^1, but not true for statements like "If f is in L^1(R), then lim_h->0+ 1/h * int_(x,x+h) f = f(x) for a.e. x." Huh? If f is in L^1 the metric space, then the sentence in quotes is nonsense; the integrals of f are well defined, but the values of f are not.

Well yes, but this applies to all sorts of statements about
L^1 "functions"; you could just as well say that "f(x) = 0 ae"
was nonsense.

 Quote: My point is that the notation L^1 is used in two ways: in the sense you mentioned and as as a space of actual functions.

It's certainly true that the notation is used in those two
senses. But I don't think it's true that whenever we see
a statement of the sort you quoted then we must conclude
equivalence classes. Careful authors define L^1 to be
a space of equivalence classes and then say that if
f is in L^1 then what you said holds. Instead of saying
that such a statement means we're talking about actual
functions one can (and at least one one does<g>) take
the point of view that if f is one of those equivalence
clases then "f(x) has property P for almost all x"
means that any representative of x has property P.

If you see a sentence that says something about
f(x) for _every_ x then the author must be talking
about actual functions, but it seems to me that the
whole point to the fact that statements about
f(x) for f in L^1 always tend to include the
qualifier "ae" is that if you include the "ae"
then the meaning of the statement is well-defined
for f equal to an equivalence class.

************************

David C. Ullrich
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Mon Jul 17, 2006 4:21 am    Post subject: Re: The Fundamental Theorem of Calculus

In article <79bkb25vrg4rh4cqbc4f17sa8i3g55b5n3@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:

 Quote: On Sat, 15 Jul 2006 11:42:13 -0700, The World Wide Wade waderameyxiii@comcast.remove13.net> wrote: A few comments on that. First, elements of L^1 are not actually functions, they're equivalence classes of functions where two functions are identified if they agree almost everywhere. True if you're in the metric space L^1, but not true for statements like "If f is in L^1(R), then lim_h->0+ 1/h * int_(x,x+h) f = f(x) for a.e. x." Huh? If f is in L^1 the metric space, then the sentence in quotes is nonsense; the integrals of f are well defined, but the values of f are not. Well yes, but this applies to all sorts of statements about L^1 "functions"; you could just as well say that "f(x) = 0 ae" was nonsense.

Yes, that would be nonsense too.

 Quote: My point is that the notation L^1 is used in two ways: in the sense you mentioned and as as a space of actual functions. It's certainly true that the notation is used in those two senses.

Well then, writing "First, elements of L^1 are not actually
functions" seems a little misleading to me.

 Quote: But I don't think it's true that whenever we see a statement of the sort you quoted then we must conclude that we're talking about actual functions instead of equivalence classes. Careful authors define L^1 to be a space of equivalence classes

Not so fast. Rudin, for example, defines L^p to be a space of
measurable functions - actual functions. Later he notes that if
L^p is to be regarded as a metric space, we need to invoke
equivalence classes. But he certainly never disavows the original
definition, for good reasons IMHO.

 Quote: and then say that if f is in L^1 then what you said holds. Instead of saying that such a statement means we're talking about actual functions one can (and at least one one does) take the point of view that if f is one of those equivalence clases then "f(x) has property P for almost all x" means that any representative of x has property P.

That solves the particular problem I brought up, but very few
authors do this. Rather, they seem to allow for the dual
interpretation of L^p that I mentioned. And this seems like the
best solution to me. How else are you going to deal with
something like the Lebesgue set of f in L^1(R^k)? (See Rudin,
Theorem 8.8 for example.)
David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250

Posted: Mon Jul 17, 2006 1:05 pm    Post subject: Re: The Fundamental Theorem of Calculus

On Sun, 16 Jul 2006 21:21:26 -0700, The World Wide Wade

 Quote: In article <79bkb25vrg4rh4cqbc4f17sa8i3g55b5n3@4ax.com>, David C. Ullrich wrote: On Sat, 15 Jul 2006 11:42:13 -0700, The World Wide Wade waderameyxiii@comcast.remove13.net> wrote: A few comments on that. First, elements of L^1 are not actually functions, they're equivalence classes of functions where two functions are identified if they agree almost everywhere. True if you're in the metric space L^1, but not true for statements like "If f is in L^1(R), then lim_h->0+ 1/h * int_(x,x+h) f = f(x) for a.e. x." Huh? If f is in L^1 the metric space, then the sentence in quotes is nonsense; the integrals of f are well defined, but the values of f are not. Well yes, but this applies to all sorts of statements about L^1 "functions"; you could just as well say that "f(x) = 0 ae" was nonsense. Yes, that would be nonsense too. My point is that the notation L^1 is used in two ways: in the sense you mentioned and as as a space of actual functions. It's certainly true that the notation is used in those two senses. Well then, writing "First, elements of L^1 are not actually functions" seems a little misleading to me. But I don't think it's true that whenever we see a statement of the sort you quoted then we must conclude that we're talking about actual functions instead of equivalence classes. Careful authors define L^1 to be a space of equivalence classes Not so fast. Rudin, for example, defines L^p to be a space of measurable functions - actual functions. Later he notes that if L^p is to be regarded as a metric space, we need to invoke equivalence classes. But he certainly never disavows the original definition,

.... ohmygod you're right. I'd be struck dead immediately if I
said I didn't approve, because this makes the meaning of the
notation dependent on our state of mind. So I won't say that.

Folland otoh labels the definition of L^1 as a space of
functions "provisional" and then half a page later says
he's going to _redefine_ the meaning of the notation.

There do exist books that have different notations
for the two objects, which seems to me like the best
course, the only problem being that then one of the
notations has to be somewhat nonstandard.

But if Rudin agrees with you then I certainly can't
claim that your convention is nonstandard.

 Quote: for good reasons IMHO. and then say that if f is in L^1 then what you said holds. Instead of saying that such a statement means we're talking about actual functions one can (and at least one one does) take the point of view that if f is one of those equivalence clases then "f(x) has property P for almost all x" means that any representative of x has property P. That solves the particular problem I brought up, but very few authors do this. Rather, they seem to allow for the dual interpretation of L^p that I mentioned. And this seems like the best solution to me. How else are you going to deal with something like the Lebesgue set of f in L^1(R^k)? (See Rudin, Theorem 8.8 for example.)

Well, the way _I'd_ define the Lebesgue set would be
as an element of the algebra of measurable sets modulo
null sets (and then of course I'd almost always ignore
this and speak of it as thought it were a set.)

************************

David C. Ullrich

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