proof: cauchy-sequence

• Dec 25th 2008, 07:59 AM
Rapha
proof: cauchy-sequence
Hi! (Merry Christmas and happy holidays!! I hope you guys have a good time)

Exercise

Proof: (F, ||-||_H) is a Banach space, where

$\displaystyle F := \{f \in L^2(\mathbb{R}) : \int^\infty_{-\infty} (1+x^2)|\hat{f}(x)|^2 dx < \infty \}$
($\displaystyle \hat{f}$ fourier transform)

and $\displaystyle ||-||_H := (\int^\infty_{-\infty}(1+x^2)* \overline{\hat{f}} *\hat{f} dx)^{0.5}$

Definition
If the norm on F is complete (that is, any Cauchy sequence in F is convergent), then F is called a Banach space

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According to the Cauchy sequence/norm I'm at a loss what to do.

Best regards,
Rapha.
• Dec 26th 2008, 12:00 AM
Opalg
Quote:

Originally Posted by Rapha
Exercise

Proof: (F, ||-||_H) is a Banach space, where

$\displaystyle F := \{f \in L^2(\mathbb{R}) : \int^\infty_{-\infty} (1+x^2)|\hat{f}(x)|^2 dx < \infty \}$
($\displaystyle \hat{f}$ fourier transform)

and $\displaystyle ||-||_H := (\int^\infty_{-\infty}(1+x^2)* \overline{\hat{f}} *\hat{f} dx)^{0.5}$

Let U be the map from F to $\displaystyle L^2(\mathbb{R})$ given by $\displaystyle (Uf)(x) = \sqrt{1+x^2}\hat{f}(x)$. Then $\displaystyle \|U(f)\|_2 = \|f\|_H$. The map U is linear, and it follows that F, with the norm H, is a normed linear space. To show that it is a Banach space, you need to show that U is surjective. The completeness of F then follows from that of $\displaystyle L^2(\mathbb{R})$.