Why is $L^p([-\pi,\pi])$ isomorphic to $L^p(\{\Vert z \Vert_2 = 1\})$?
On the contrary, why is $C([-\pi,\pi])$ not isomorphic to $C(\{\Vert z \Vert_2 = 1\})$?
Why is $L^p([-\pi,\pi])$ isomorphic to $L^p(\{\Vert z \Vert_2 = 1\})$ and $C([-\pi,\pi])$ not isomorphic to $C(\{\Vert z \Vert_2 = 1\})$?
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functional-analysis
functions
lp-spaces
1 Answers
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Let $f\in L^p([-\pi,\pi))$, then $g(\exp(ix)) = f(x)$ defines a function on the torus $\Bbb T$. It is easy to see that it is an $L^p$ function:
$$\int_{-\pi}^{\pi} |f(x)|^2\,dx = \int_{-\pi}^{\pi} |g(\exp(ix)|^2\,dx.$$
Here $x$ goes from being the real line measure to the angular measure. $L^p([-\pi,\pi))$ is then isomorphic to $L^p([-\pi,\pi])$ since we can just tack on a point without changing anything since our measure is not discrete.
This does not extend to continuous functions. We would have that $C([-\pi,\pi))$ is isomorphic to $C(\Bbb T)$ much in the same way with $L^p$. If we wanted to make the isomorphism you are suggesting, we would need to find a continuous function (composition of continuous functions is continuous) from $\Bbb T$ to $[-\pi,\pi]$ and vice versa. This doesn't work because continuous functions preserve connectedness. If you remove a point from $\Bbb T$, it is still connected. This is not the case for an interval.
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0In the first part of your answer, why is it still an $L^p$ function? In the second part, actually the image of a simply-connected set under a continuous function need not be simply connected. – 2017-01-30
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0Why is the composition of an $L^p$ function and a continuous function again $L^p$? – 2017-01-30
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0Let me make some changes. What I said isn't quite right. I have a headache. – 2017-01-30
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0I meant to say connected. My apologies. – 2017-01-30
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0Why are those integrals equal? – 2017-01-30
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0By definition of $g$. It's a bijective mapping as well so there's no extra work to do. – 2017-01-30