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Let $L^2(\mu)$ and $L^2(\nu)$ with respect to two different positive measures, then they are two Banach spaces. I'm considering whether the space $$L^2(\mu)+L^2(\nu)$$ is still a Banach space?

e.g. $\mu$ be Lebesgue measure, $d\nu=ln(1+|x|)d\mu$, my idea is that since both $L^2(\mu)$ and $L^2(\nu)$ are continuous embedded to the measurable functions space $\mathcal M$, it's done.

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    $\begingroup$ What is your definition of $L^2(\mu)+L^2(\nu)$? $\endgroup$ – Disintegrating By Parts Dec 22 '18 at 20:30
  • $\begingroup$ Are $\mu, \nu$ defined on the same sigma algebra? $\endgroup$ – Alex Vong Dec 23 '18 at 20:43
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If you have two Banach spaces $X,Y$ both continuously embedded in a Hausdorff topological vector space $Z$, you can endow the sum $X+Y$ with the norm $\|z\|=\inf\{\|x\|_X+\|y\|_Y: z=x+y\}$. Then $X+Y$ is a quotient of the Banach space $X\times Y$ with respect to the subspace $X\cap Y$ and hence itself a Banach space.

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  • $\begingroup$ Thank you! But for my case, what's the space $Z$? $\endgroup$ – Yixuan Zhang Jan 3 '19 at 8:31
  • $\begingroup$ You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $\mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)\mapsto f-g$ that the kernel $X\cap Y$ is closed). $\endgroup$ – Jochen Jan 3 '19 at 13:33
  • $\begingroup$ well, thank you very much! $\endgroup$ – Yixuan Zhang Jan 5 '19 at 13:18

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