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Math Help - Dimension of a normed space

  1. #1
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    Dimension of a normed space

    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




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  2. #2
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    Quote Originally Posted by Rapha View Post
     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.
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  3. #3
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    Hello opalg

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

    Thank you,

    best wishes
    Rapha
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  4. #4
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    f is periodic function which compose infinite cos & sin . It's demension is infinite.
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