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Apr 12th 2011, 04:32 PM
surjective

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Hey,

I have a map:

$T: L^{p}(-2,2) \to L^{p}(-2,2)$ , $(Tf)(x)=xf(x)$

$T$ maps from $L^{p}(-2,2)$ to $L^{p}(-2,2)$ because:

$\int_{-2}^{2}|(Tf)(x)|^{p}dx=\int_{-2}^{2}|xf(x)|^{p}dx=\int_{-2}^{2}|x|^{p}|f(x)|^{p}dx=?$

How to continue so that it shows that $\int_{-\infty}^{\infty}|f(x)|^{p}dx < \infty$
Apr 12th 2011, 11:57 PM
Opalg

$\int_{-2}^{2}|(Tf)(x)|^{p}\,dx=\int_{-2}^{2}|xf(x)|^{p}\,dx=\int_{-2}^{2}|x|^{p}|f(x)|^{p}\,dx \leqslant \int_{-2}^{2}2^{p}|f(x)|^{p}\,dx = \ldots$ .
Apr 13th 2011, 06:41 AM
surjective

This is because the lrgest value x may assume is 2. Right?
Apr 13th 2011, 08:27 AM
Opalg

Quote:

Originally Posted by surjective

This is because the lrgest value x may assume is 2. Right?

Yes. (Nod)

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