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Apr 27, 2020 at 18:39 comment added ABIM Just as a general point, your space is just equal to $L^{1/2}([0,\infty))$ when you change measure to $\nu$, defined by the Radon-Nikodym derivative $\frac{d\nu}{dm}(x) = \sum_{n=1}^{\infty} \frac1{2^n} I_{[n,n+1]}(x)$ (here $m$ is the Lebesgue measure).
Apr 27, 2020 at 15:16 history edited ABIM CC BY-SA 4.0
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Apr 27, 2020 at 14:07 history edited ABIM CC BY-SA 4.0
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Apr 27, 2020 at 12:59 history edited YCor CC BY-SA 4.0
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Apr 27, 2020 at 12:55 history asked ABIM CC BY-SA 4.0