Re: fourier series of f ' relative to f

On 5 huhti, 20:45, W^3 <aderamey.a...@xxxxxxxxxxx> wrote:
In article <9c4ht4pd2op8ib24gbge1mdkslkihj0...@xxxxxxx>,
 David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:



On Sat, 4 Apr 2009 05:24:44 -0700 (PDT), Gc <Gcut...@xxxxxxxxxxx>
wrote:

On 1 huhti, 17:33, G Patel <gaya.pa...@xxxxxxxxx> wrote:
What is the theorem that can allow me to justify moving
differentiation operator inside infinite sums (or infinite limits)?

I think you need absolute convergence of the original series.

No, you need uniform convergence of the original series, and
_also_ uniform convergence of the series of derivatives.

If your
periodic function is continuosly differentiable, in other words belong
to C^1, then you have it.

Yes, C^1 implies the Fourier series for f converges absolutely.
But that's not enough to say the derivative of the sum is the
sum of the derivatives. C^1 does _not_ imply that the
differentiated series converges uniformly (precisely because
there exist continuous functions with divergent Fourier
series; take f so that f' is continuous with a divergent
Fourier series.)

A condition known as "C^(1 + epsilon)" is enough. But
(i) that doesn't answer the OP's question, and (ii)
the actual solution is very simple, just using the
definition of the Fourier coefficients instead of
trying to differentiate the series.

The basic idea is simple, but I don't see how to follow through
without measure theory.

Maybe you forgot that the derivative f` was bounded. By mean value
theorem for any (a,b) on [-pii,pii], there is a c on (a,b) such that
f`(c)(b - a) = f(b) - f(a) so when you take sup f`(c) on [-pii,pii]
you obtain bound sup f`(c)(b - a) for all of the variations.


I think the problem was either misstated by
the OP or badly chosen by its author. If the author intended a
"classical" proof using the Riemann integral, then it can't be done as
stated. (Maybe the author intended piecewise C^1.) If an appeal to the
Lebesgue integral was intended, then why not just assume f is
absolutely continuous? You get a stronger and more natural result -
and the "piecewise" business can be left out.

David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)



.

Relevant Pages

  • Re: fourier series of f relative to f
    ... differentiation operator inside infinite sums? ... _also_ uniform convergence of the series of derivatives. ... periodic function is continuosly differentiable, ... But that's not enough to say the derivative of the sum is the ...
    (sci.math)
  • Re: fourier series of f relative to f
    ... differentiation operator inside infinite sums? ... _also_ uniform convergence of the series of derivatives. ... periodic function is continuosly differentiable, ... But that's not enough to say the derivative of the sum is the ...
    (sci.math)
  • Re: fourier series of f relative to f
    ... differentiation operator inside infinite sums? ... _also_ uniform convergence of the series of derivatives. ... periodic function is continuosly differentiable, ... But that's not enough to say the derivative of the sum is the ...
    (sci.math)
  • Re: fourier series of f relative to f
    ... differentiation operator inside infinite sums? ... _also_ uniform convergence of the series of derivatives. ... periodic function is continuosly differentiable, ... But that's not enough to say the derivative of the sum is the ...
    (sci.math)
  • Re: fourier series of f relative to f
    ... differentiation operator inside infinite sums? ... _also_ uniform convergence of the series of derivatives. ... periodic function is continuosly differentiable, ... But that's not enough to say the derivative of the sum is the ...
    (sci.math)