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A property of inner product of functions... Does it have a name?

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Golabi Doon

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Mar 22, 2011, 8:24:56 PM3/22/11
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Hello everyone,

I am writing a report in which I am extensively using the following
property. Given a function f(x) from [-pi,pi] to R such that it has a
fourier series representation.

Then the following inner product inequality holds:
\int_{-pi}^{pi} f_{n}(x) f_{n+2}(x) dx <= 0 ,
where n=1,2,3... and f_{n} is the n'th derivative of f.

This argument can be easily proved using the Fourier representation
f(x)=a_0 + \sum_{k=1}^\infty a_k cos(kx) + b_k sin(kx). For example,
when n=1 we have:
f_{1}(x)=\sum_{k=1}^\infty (-k a_k) sin(kx) + (k b_k) cos(kx)
f_{3}(x)=\sum_{k=1}^\infty (k^3 a_k) sin(kx) + (-k^3 b_k) cos(kx)
Thus, int_{-pi}^{pi} f_{1}(x) f_{3}(x) dx = \sum_{k=1}^\infty (-k^4
a_k^2) + (-k^4 b_k^2) <= 0 .

I would like to refer to this property in my article instead of
writing its proof, because I believe it is so elementary that it must
be listed somewhere in articles/books. Does any one know a name or a
reference for citing this property?

Regards

Golabi

David C. Ullrich

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Mar 24, 2011, 10:58:49 AM3/24/11
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I doubt that such a simple thing has a name.

You should note however that first "has a fourier
series representation" is much too vague - a function
can have such a representation valid in this sense
or in that sense.

And you should note as well that whatever you mean
by "has a fourier series representation" it's not
sufficient to make the argument above valid - you
need assumptions on the differentiability of f
(in any of various senses).

>Regards
>
>Golabi

Rob Johnson

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Mar 24, 2011, 4:09:14 PM3/24/11
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In article <73e0583e-662c-4d02...@s18g2000prg.googlegroups.com>,

You don't need to use Fourier series to prove this inequality.

For any periodic function that has two derivatives, one continuous,
on [-pi,pi], we have, using integration by parts:

|\pi
| f(x) f''(x) dx
\|-pi

|\pi
= | f(x) df'(x)
\|-pi

| pi |\pi
= f(x) f'(x) | - | f'(x) df(x)
|-pi \|-pi

|\pi 2
= - | f'(x) dx
\|-pi

<= 0

On this inner product space, derivation is skew-adjoint.

Rob Johnson <r...@trash.whim.org>
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