How Does the Schwarz Inequality Apply to Fourier Coefficients in C[-pi, pi]?

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Discussion Overview

The discussion centers on the application of the Schwarz inequality to Fourier coefficients of continuous functions defined on the interval [-π, π]. Participants explore the implications of the inequality in the context of calculating bounds for Fourier coefficients, addressing both the mathematical formulation and specific calculations involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the Fourier coefficient |h(n)| can be bounded by the expression |h(n)| ≤ (1/2π)(||f||·||e^{int}||), leading to a specific calculation involving ||e^{int}||.
  • One participant agrees with the initial claim but identifies a calculation mistake, suggesting that |h(n)| ≤ √(2π) ||f|| instead.
  • Another participant challenges the correction, arguing that the correct interpretation should yield |h(n)| ≤ (2π)^(-0.5) ||f||, rather than the proposed √(2π) ||f||.
  • A later reply acknowledges a similar mistake in their own calculations, suggesting a revised bound of |h(n)| ≤ (1/√(2π)) ||f||.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct interpretation and calculation of the bounds for |h(n)|, with multiple competing views on the application of the Schwarz inequality and the resulting expressions.

Contextual Notes

Limitations in the discussion include unresolved mathematical steps and differing interpretations of the calculations involving the norms and the Schwarz inequality.

Poirot1
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Let C[-pi,pi] be the set of continuous function from [-pi,pi] to C. Endow this with usual inner product (<f,g>= integral from -pi to pi of f multiplied by g conjugate, and let ||.|| be the corresponding norm).
Let h(n) be Fourier coefficient of fNow, |h(n)|<_ 1/2pi( ||f||.||e^int||) by schwarz inequaity

=1/pi . 1/ 2^(0.5) ||f|| since ||e^int|| =(pi +pi)^0.5

Do you agree?
 
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Re: use of schwarz inequality

Poirot said:
Let C[-pi,pi] be the set of continuous function from [-pi,pi] to C. Endow this with usual inner product (<f,g>= integral from -pi to pi of f multiplied by g conjugate, and let ||.|| be the corresponding norm).
Let h(n) be Fourier coefficient of fNow, |h(n)|<_ 1/2pi( ||f||.||e^int||) by schwarz inequaity

=1/pi . 1/ 2^(0.5) ||f|| since ||e^int|| =(pi +pi)^0.5

Do you agree?

Hi Poirot!

I agree, except for a small calculation mistake.

$$|h(n)| \le \frac 1{2\pi} (||f|| \cdot ||e^{int}||) = \frac 1{2\pi} (||f|| \cdot \sqrt{2\pi}) = \sqrt{2\pi}||f||$$
 
Re: use of schwarz inequality

I like Serena said:
Hi Poirot!

I agree, except for a small calculation mistake.

Thanks for replying but I don't agree. We have (2pi)^0.5 multiplied by 1/(2pi). This is (2pi)^(-0.5), and not (2pi)^(0.5) as you suggest.
 
Re: use of schwarz inequality

Poirot said:
Thanks for replying but I don't agree. We have (2pi)^0.5 multiplied by 1/(2pi). This is (2pi)^(-0.5), and not (2pi)^(0.5) as you suggest.

Ah, my bad - now I've made a similar mistake as well.

So it should be
$$|h(n)| \le \frac 1 {\sqrt{2\pi}} ||f||$$
instead of the
$$|h(n)| \le \frac 1 \pi \cdot \frac 1 {2^{0.5}} ||f||$$
that you had.
 

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