| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Let f be a differentiable function that maps [0,1] into the complex
numbers. Moreoever, the derivative of f is continuous on [0,1].
Suppose
/1
(
) f(x) dx = 0 (1)
/ 0
Show that there exists a real number p > 0 such that
/1 /1
( 2 ( 2
) |f(x)| dx <= p ) |f'(x)| dx (2)
/ 0 / 0
for all f that meets the condition described above. In other word,
the number p has to work for all functions that satisfy the hypothesis
(namely all the functions that are continuously differentiable on
[0,1] and (1) holds).
Note f' means the derivative of f.
Eugene
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1040.1 | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sat Mar 25 1989 14:39 | 50 | |
The Riemann integral of a functional continuous over [a,b]
always exists, and here f and f' are both continuous.
Expand f into it's Fourier expansion
2 pi n i x
f(x) = sum(-oo < n < +oo) c e
n
-2 pi n i x
where each c is the integral over [0,1] of f(x)e dx
n
Also, f' has a Fourier expansion into the series with
coefficients d for -oo < n < +oo.
n
The proof will consist of showing (or looking up):
2 2
1) integral of |f(x)| dx over [0,1] = sum |c |
n
[I think that's called Parseval's identity.]
2) same as (1) but for f' and the d
n
3) The d can be obtained by differentiating the Fourier
n
series of f(x) term by term.
[This is the step that requires careful justification.]
4) c is zero because the integral of f(x) dx over [0,1] is 0.
0
5) Now compare the two integrals by comparing the
Parseval's sums of the series coefficients. Each term
in the "d" series for f' is 2 pi |n| times the
corresponding term in the "c" series for f.
2 2
Therefore the integral of |f| <= p * integral of |f'|
whenever p >= 1/(2 pi).
6) Let f(x) = sin(2 pi x). Then no value of p < 1/(2 pi)
will work in the inequality. Note that for this f the
coefficients c are all zero for |n| > 1.
n
Dan
| |||||
| 1040.2 | HPSTEK::XIA | Sat Mar 25 1989 19:06 | 13 | ||
re -1
Dan,
You got it right! I think you should go and get an advanced degree
in math :-). Now you can do my 1040A :-). As for the problem in note
1039, you can assume the function is continuous too (that should make the
problem less complicated). Well, at least do it for the continuous
function first....
Finally, congratulations. This ain't no trivial problem.
Eugene
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