How does one prove that $L^p(\mathbb{R})$ is isomterically isomorphic to $L^p((0,1))$? If anyone can give some hints it will be greatly useful! We assume that both spaces use the Lebesgue measure.
$L^p(\mathbb{R})$ is isomterically isomorphic to $L^p((0,1))$
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functional-analysis
lp-spaces
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2Let $\varphi \colon (0,1) \to \mathbb{R}$ be a differentiable bijection. You know that for nice enough $F$ you have $$\int_{\varphi(a)}^{\varphi(b)} F(x)\,dx = \int_a^b F(\varphi(t))\cdot \varphi'(t)\,dt.$$ Does that give you an idea? – 2017-01-11
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0I am really sorry. I do not understand your clue! – 2017-01-11
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0For $f \in L^p(\mathbb{R})$, let $F(x) = \lvert f(x)\rvert^p$. Then on the left hand side, you get $\lVert f\rVert_{L^p(\mathbb{R})}^p$ when $a = 0,\, b = 1$ (assuming $\varphi$ is increasing). On the right, you then want to have $\lVert T(f)\rVert_{L^p((0,1))}^p$, where $T$ is the isometric isomorphism. – 2017-01-11
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0Yes I got it now. I believe we can create the bijection using the arctan function. Thanks! – 2017-01-11