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

[Poirot1]

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 coefficent 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?

 

[I like Serena]

Re: use of schwarz inequality

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 coefficent 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.

$$|h(n)| \le \frac 1{2\pi} (||f|| \cdot ||e^{int}||) = \frac 1{2\pi} (||f|| \cdot \sqrt{2\pi}) = \sqrt{2\pi}||f||$$

 

[Poirot1]

Re: use of schwarz inequality

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.

 

[I like Serena]

Re: use of schwarz inequality

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.