Dimension of a normed space

**Rapha** · January 3rd 2009, 11:42 PM

Hello!

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

Show that $dim(F) < \infty$

I really don't know how to do this. Usually you find n basis vectors of X => dim(X) = n. But I find it pretty hard, anyone got some ideas? Any help would be much appreciated.

Thank you,
Rapha

**Opalg** · January 4th 2009, 03:03 AM

Originally Posted by Rapha

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

Show that $dim(F) < \infty$

I really don't know how to do this. Usually you find n basis vectors of X => dim(X) = n. But I find it pretty hard, anyone got some ideas? Any help would be much appreciated.

The space F is not finite-dimensional. In fact, let $G := \{g \in L^2(\mathbb{R}) : \int^\infty_{-\infty} (1+x^2)|g(x)|^2 dx < \infty \}$ . Then G is infinite-dimensional. (For example, it will contain all the functions in $L^2(\mathbb{R})$ that have compact support.) But F is the image of G under the inverse Fourier transform, so F is also infinite-dimensional.

**Rapha** · January 4th 2009, 04:10 AM

Hello opalg

thank you so much for replying to such a difficult problem!

Thank you,

best wishes
Rapha

**math2009** · January 4th 2009, 07:49 PM

$f$ is periodic function which compose infinite cos & sin . It's demension is infinite.

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