| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1039.1 | proof | AIRPRT::GRIER | mjg's holistic computing agency | Sun Mar 19 1989 22:56 | 8 |
| Assume the hypothesis.
Obvious. QED
-mjg
:-)
|
| 1039.2 | Seriously, tho' | AIRPRT::GRIER | mjg's holistic computing agency | Mon Mar 20 1989 01:35 | 7 |
| I know how to do this in the case where p=1, I'm not sure how to generalize.
It seems "obvious", but most of the interesting stuff in mathematics isn't
obvious, now, is it? (And then some "obvious" things aren't true...)
-mjg (who feels dumb after two full
semesters of Complex Analysis and
is stumped at the moment)
|
| 1039.3 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Mon Mar 20 1989 04:16 | 13 |
| Are we allowed to assume that if f is integrable on [0,1]
then |f(x)|^p will also be integrable on [0,1]? Or do we
have to prove that, too? (For real p >= 1, of course.)
Riemann or Lebesgue integration? :-)
I suppose a proof using Riemann integration would be done by
proving a version of the theorem for finite sums, and then
applying that to approximations of the integrals by finite
sums.
Dan
|
| 1039.4 | | HPSTEK::XIA | | Mon Mar 20 1989 14:06 | 5 |
| re -1
Make it Riemann integrable.
Eugene
|
| 1039.5 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Mon Mar 20 1989 23:12 | 34 |
| Okay. The usual definition of Riemann integration applies
to a bounded function. Lebesgue integration theory shows
that the Riemann integral of a bounded function will exist
if and only if the function is continuous almost everywhere,
i.e., except on a set of Lebesgue measure zero. So if f is
Riemann integrable over [0,1] and is bounded, then it must
be continuous almost everywhere, and then |f|^p is bounded
and continuous almost everywhere, and so its integral
exists.
If f is unbounded, then its Riemann integral would be
defined as a limit as the bound goes to infinity of the
integral of a "truncated" f that was bounded. It's not
clear to me that the limit would still exist after raising
|f| to the p power (p >= 1), which would make the "large"
parts of f even larger.
[I'm obviously stalling until I can work out an approach to
proving the inequality.] :-)
Hmm. Is
/1 /1
( p ( p
) |f(x)| dx - | ) f(x) dx |
/ 0 / 0
a differential function of p, and if so is the derivative
always nonnegative? Those two conditions would suffice
because the inequality is true for p=1 (it's not too hard to
show that).
Dan
|
| 1039.6 | Solution :-) :-) | HPSTEK::XIA | | Wed May 24 1989 02:32 | 30 |
| By Holder's inequality,
/1 /1 /1
( p 1/p ( q 1/q (
[ ) |f(x)| dx ] * [ ) 1 dx ] >= | ) f(x) * 1 dx|
/ 0 / 0 / 0
Where 1/p + 1/q = 1 and p >= 0, q >= 0.
So we have:
/1 /1
( p 1/p (
[ ) |f(x)| dx ] >= | ) f(x) dx|
/ 0 / 0
Then it is immediate that:
/1 /1
( p ( p
) |f(x)| dx >= | ) f(x) dx|
/ 0 / 0
Eugene
|