grubb@math.niu.edu
Apr7-04, 08:45 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>robert bristow-johnson <rbj@surfglobal.net> wrote in message news:<BC91EE14.A129%rbj@surfglobal.net>...\n> In article 551fc6d2.0403311105.53306b49@posting.google.com,\n > grubb@math.niu.edu at grubb@math.niu.edu wrote on 04/01/2004 10:28:\n>\n> > Charles Francis <charles@clef.demon.co.uk> wrote in message\n> > news:<ovT3vtwvM8ZAFw1Z@clef.demon.co.uk>...\n> >> In article <BC80EFBD.9A0A%rbj@surfglobal.net>, robert bristow-johnson\n> >> <rbj@surfglobal.net> writes\n> >>>\n> >>> +a\n> >>> integral{ d(t) dt} = 1\n> >>> -a\n> >>> a > 0\n> >>> +a\n> >>> integral{ 0 dt} = 0\n> >>> -a\n> >>>\n> >>> and 0 is usually not the same value as 1 so Fatou\'s Lemma is violate\n> >>\n> >>\n> >> d(0) is undefined, so Fatou\'s lemma does not apply\n> >\n> > This is not the reason Fatou doesn\'t apply. In fact, all that Fatou\n> > requires is a function that is defined almost everywhere. The problem\n> > is that the delta \'function\' is not a function, even one defined\n> > only almost everywhere.\n> >\n> > Furthermore, what usually goes as Fatou\'s lemma goes as follows:\n>\n> hey guys, it has already been pointed out by Robert Egri and acknowledged by\n> me that i misascribed this to Fatou\'s Lemma. it was a note in my old class\n> notes from a class in Real Analysis that i took in the late 70s. in the\n> book ("Real Analysis" H.L.Royden), this "proposition" is first stated on p.\n> 80, one section before Fatou. sorry for the misattribution.\n\nSorry I didn\'t see that. I didn\'t mean to harp.\n\n>\n> just FWIW, i think the explanation that has helped me the most (none has\n> completely cut it, but i recognize that the problem probably exists more\n> with my inability to get my mind wrapped around this) is that, perhaps, for\n> a function candidate to qualify as a "true" function, it cannot be allowed\n> to "remember" how it may have been defined from a limit of real functions.\n> it can only be defined in terms of how it maps real numbers to other real\n> numbers.\n>\n> i.e. there are a lot (an infinite number) of possibilities for some f(x) = 0\n> for |x|>0 and infinite (or undefined) for x=0. for f(x) to be a normal real\n> function (as defined by the powers that be), we cannot add to its definition\n> any more than the function mapping (as stated above). we cannot add to its\n> definition that it was a limit of some other functions that all had an area\n> of 1. only the basic real-to-real mapping is allowed.\n>\n\nNo. This is not what is going on. There are not an infinite number of\nways a function can be infinite at a point. A function is a function\nis a function. It takes in some things from some set and spits out\nthings from another (or possibly the same) set. The \'dirac\' delta is\nnot an extended real-valued function on the real line. (In other\nwords,\nit does not take real numbers as arguments and spit out\nreal-or-infinite\nvalues.) It is a distribution.\n\nNow, what is a distribution? It is a function (some would prefer to\ncall it an operator) that takes a function and spits out a real\nnumber.\nThere are some continuity requirements concerning derivatives and all\nand some details about which functions can be operated on, i.e. \'the\ntest\nfunctions\', but that is the essence. Oh, distributions are also\nsupposed\nto be linear, i.e. if we operate on a linear combination of functions,\nthe result is the linear combination of the results of operating on\nthe\nindividual functions.\n\nThus, if L is a distribution, we want L(af+bg)=aL(f) +bL(g) where\nf and g are test functions and a,b are real (or complex) numbers.\nIt\'s important to know that the test functions are all infinitely\ndifferentiable functions, and any derivative of a test function\nis again a test function. Hence, if f is a test function, we can\nevaluate f^(n)(x), the n^th derivative of f at any point x. We also\nrequire that test functions be \'rapidly decreasing\' at infinity.\n\nPerhaps some examples are in order. First, let k be any integrable\nfunction. We can define a distribution by setting L_k (f) to be\nthe integral of f(x)k(x) from -\\infty to \\infty. This is a linear\noperator that takes test functions to numbers and has the required\ncontinuity. *By convention*, this distribution and the function k\nare identified, even though they are different creatures. By\nproperties\nof the integral, two k\'s that differ only on a set of measure 0\nare identified as distributions.\n\nThe Dirac Delta is the distribution that takes a test function f and\nspits out the value f(0). That\'s it. Now, this distribution cannot\nbe obtained from any k as in the previous paragraph, so the delta\nis not a function (by the previous abuse of terminology). Another\ndistribution that cannot be obtained from a k is the \'derivative\nof the delta\', where L(f)=-f\'(0).\n\nUnfortunately, people like to confuse matters and think of the delta\nas a function and think of the integral of some delta(x)f(x) as giving\nf(0). Not strictly true, but good enough if you are careful not to\ngive it too much credence.\n\nIt is possible to write the action of many distributions by finding a\nsequence k_n of integrable functions and writing L(f) as the limit of\nL_(k_n) (f). For the delta, gaussians, and box functions, etc. can be\nused for such sequences \'converging\' to the delta. This is not\npointwise\nconvergence of functions!!! To be strict, it is a \'weak\' convergence\nof linear maps.\n\nIt should be pointed out that there are functions that converge to the\nderivative of the delta as in the previous paragraph. Unlike for the\ndelta, though, these functions can be chosen to converge pointwise\n*everywhere* to the zero function (even at x=0).\n\n---Dan Grubb\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>robert bristow-johnson <rbj@surfglobal.net> wrote in message news:<BC91EE14.A129%rbj@surfglobal.net>...
> In article 551fc6d2.0403311105.53306b49@posting.google.com,
> grubb@math.niu.edu at grubb@math.niu.edu wrote on 04/01/2004 10:28:
>
> > Charles Francis <charles@clef.demon.co.uk> wrote in message
> > news:<ovT3vtwvM8ZAFw1Z@clef.demon.co.uk>...
> >> In article <BC80EFBD.9A0A%rbj@surfglobal.net>, robert bristow-johnson
> >> <rbj@surfglobal.net> writes
> >>>
> >>> +a
> >>> integral{ d(t) dt} = 1
> >>> -a
> >>> a >
> >>> +a
> >>> integral{ dt} =
> >>> -a
> >>>
> >>> and is usually not the same value as 1 so Fatou's Lemma is violate
> >>
> >>
> >> d(0) is undefined, so Fatou's lemma does not apply
> >
> > This is not the reason Fatou doesn't apply. In fact, all that Fatou
> > requires is a function that is defined almost everywhere. The problem
> > is that the \delta 'function' is not a function, even one defined
> > only almost everywhere.
> >
> > Furthermore, what usually goes as Fatou's lemma goes as follows:
>
> hey guys, it has already been pointed out by Robert Egri and acknowledged by
> me that i misascribed this to Fatou's Lemma. it was a note in my old class
> notes from a class in Real Analysis that i took in the late 70s. in the
> book ("Real Analysis" H.L.Royden), this "proposition" is first stated on p.
> 80, one section before Fatou. sorry for the misattribution.
Sorry I didn't see that. I didn't mean to harp.
>
> just FWIW, i think the explanation that has helped me the most (none has
> completely cut it, but i recognize that the problem probably exists more
> with my inability to get my mind wrapped around this) is that, perhaps, for
> a function candidate to qualify as a "true" function, it cannot be allowed
> to "remember" how it may have been defined from a limit of real functions.
> it can only be defined in terms of how it maps real numbers to other real
> numbers.
>
> i.e. there are a lot (an infinite number) of possibilities for some f(x) =
> for |x|>0 and infinite (or undefined) for x=0. for f(x) to be a normal real
> function (as defined by the powers that be), we cannot add to its definition
> any more than the function mapping (as stated above). we cannot add to its
> definition that it was a limit of some other functions that all had an area
> of 1. only the basic real-to-real mapping is allowed.
>
No. This is not what is going on. There are not an infinite number of
ways a function can be infinite at a point. A function is a function
is a function. It takes in some things from some set and spits out
things from another (or possibly the same) set. The 'dirac' \delta is
not an extended real-valued function on the real line. (In other
words,
it does not take real numbers as arguments and spit out
real-or-infinite
values.) It is a distribution.
Now, what is a distribution? It is a function (some would prefer to
call it an operator) that takes a function and spits out a real
number.
There are some continuity requirements concerning derivatives and all
and some details about which functions can be operated on, i.e. 'the
test
functions', but that is the essence. Oh, distributions are also
supposed
to be linear, i.e. if we operate on a linear combination of functions,
the result is the linear combination of the results of operating on
the
individual functions.
Thus, if L is a distribution, we want L(af+bg)=aL(f) +bL(g) where
f and g are test functions and a,b are real (or complex) numbers.
It's important to know that the test functions are all infinitely
differentiable functions, and any derivative of a test function
is again a test function. Hence, if f is a test function, we can
evaluate f^(n)(x), the n^{th} derivative of f at any point x. We also
require that test functions be 'rapidly decreasing' at infinity.
Perhaps some examples are in order. First, let k be any integrable
function. We can define a distribution by setting L_k (f) to be
the integral of f(x)k(x) from -\infty to \infty. This is a linear
operator that takes test functions to numbers and has the required
continuity. *By convention*, this distribution and the function k
are identified, even though they are different creatures. By
properties
of the integral, two k's that differ only on a set of measure
are identified as distributions.
The Dirac \Delta is the distribution that takes a test function f and
spits out the value f(0). That's it. Now, this distribution cannot
be obtained from any k as in the previous paragraph, so the \delta
is not a function (by the previous abuse of terminology). Another
distribution that cannot be obtained from a k is the 'derivative
of the \delta', where L(f)=-f'(0).
Unfortunately, people like to confuse matters and think of the \delta
as a function and think of the integral of some \delta(x)f(x) as giving
f(0). Not strictly true, but good enough if you are careful not to
give it too much credence.
It is possible to write the action of many distributions by finding a
sequence k_n of integrable functions and writing L(f) as the limit of
L_(k_n) (f). For the \delta, gaussians, and box functions, etc. can be
used for such sequences 'converging' to the \delta. This is not
pointwise
convergence of functions!!! To be strict, it is a 'weak' convergence
of linear maps.
It should be pointed out that there are functions that converge to the
derivative of the \delta as in the previous paragraph. Unlike for the
\delta, though, these functions can be chosen to converge pointwise
*everywhere* to the zero function (even at x=0).
---Dan Grubb
> In article 551fc6d2.0403311105.53306b49@posting.google.com,
> grubb@math.niu.edu at grubb@math.niu.edu wrote on 04/01/2004 10:28:
>
> > Charles Francis <charles@clef.demon.co.uk> wrote in message
> > news:<ovT3vtwvM8ZAFw1Z@clef.demon.co.uk>...
> >> In article <BC80EFBD.9A0A%rbj@surfglobal.net>, robert bristow-johnson
> >> <rbj@surfglobal.net> writes
> >>>
> >>> +a
> >>> integral{ d(t) dt} = 1
> >>> -a
> >>> a >
> >>> +a
> >>> integral{ dt} =
> >>> -a
> >>>
> >>> and is usually not the same value as 1 so Fatou's Lemma is violate
> >>
> >>
> >> d(0) is undefined, so Fatou's lemma does not apply
> >
> > This is not the reason Fatou doesn't apply. In fact, all that Fatou
> > requires is a function that is defined almost everywhere. The problem
> > is that the \delta 'function' is not a function, even one defined
> > only almost everywhere.
> >
> > Furthermore, what usually goes as Fatou's lemma goes as follows:
>
> hey guys, it has already been pointed out by Robert Egri and acknowledged by
> me that i misascribed this to Fatou's Lemma. it was a note in my old class
> notes from a class in Real Analysis that i took in the late 70s. in the
> book ("Real Analysis" H.L.Royden), this "proposition" is first stated on p.
> 80, one section before Fatou. sorry for the misattribution.
Sorry I didn't see that. I didn't mean to harp.
>
> just FWIW, i think the explanation that has helped me the most (none has
> completely cut it, but i recognize that the problem probably exists more
> with my inability to get my mind wrapped around this) is that, perhaps, for
> a function candidate to qualify as a "true" function, it cannot be allowed
> to "remember" how it may have been defined from a limit of real functions.
> it can only be defined in terms of how it maps real numbers to other real
> numbers.
>
> i.e. there are a lot (an infinite number) of possibilities for some f(x) =
> for |x|>0 and infinite (or undefined) for x=0. for f(x) to be a normal real
> function (as defined by the powers that be), we cannot add to its definition
> any more than the function mapping (as stated above). we cannot add to its
> definition that it was a limit of some other functions that all had an area
> of 1. only the basic real-to-real mapping is allowed.
>
No. This is not what is going on. There are not an infinite number of
ways a function can be infinite at a point. A function is a function
is a function. It takes in some things from some set and spits out
things from another (or possibly the same) set. The 'dirac' \delta is
not an extended real-valued function on the real line. (In other
words,
it does not take real numbers as arguments and spit out
real-or-infinite
values.) It is a distribution.
Now, what is a distribution? It is a function (some would prefer to
call it an operator) that takes a function and spits out a real
number.
There are some continuity requirements concerning derivatives and all
and some details about which functions can be operated on, i.e. 'the
test
functions', but that is the essence. Oh, distributions are also
supposed
to be linear, i.e. if we operate on a linear combination of functions,
the result is the linear combination of the results of operating on
the
individual functions.
Thus, if L is a distribution, we want L(af+bg)=aL(f) +bL(g) where
f and g are test functions and a,b are real (or complex) numbers.
It's important to know that the test functions are all infinitely
differentiable functions, and any derivative of a test function
is again a test function. Hence, if f is a test function, we can
evaluate f^(n)(x), the n^{th} derivative of f at any point x. We also
require that test functions be 'rapidly decreasing' at infinity.
Perhaps some examples are in order. First, let k be any integrable
function. We can define a distribution by setting L_k (f) to be
the integral of f(x)k(x) from -\infty to \infty. This is a linear
operator that takes test functions to numbers and has the required
continuity. *By convention*, this distribution and the function k
are identified, even though they are different creatures. By
properties
of the integral, two k's that differ only on a set of measure
are identified as distributions.
The Dirac \Delta is the distribution that takes a test function f and
spits out the value f(0). That's it. Now, this distribution cannot
be obtained from any k as in the previous paragraph, so the \delta
is not a function (by the previous abuse of terminology). Another
distribution that cannot be obtained from a k is the 'derivative
of the \delta', where L(f)=-f'(0).
Unfortunately, people like to confuse matters and think of the \delta
as a function and think of the integral of some \delta(x)f(x) as giving
f(0). Not strictly true, but good enough if you are careful not to
give it too much credence.
It is possible to write the action of many distributions by finding a
sequence k_n of integrable functions and writing L(f) as the limit of
L_(k_n) (f). For the \delta, gaussians, and box functions, etc. can be
used for such sequences 'converging' to the \delta. This is not
pointwise
convergence of functions!!! To be strict, it is a 'weak' convergence
of linear maps.
It should be pointed out that there are functions that converge to the
derivative of the \delta as in the previous paragraph. Unlike for the
\delta, though, these functions can be chosen to converge pointwise
*everywhere* to the zero function (even at x=0).
---Dan Grubb