Let $X$ be a topological space and $\mu$ be the Borel measure on $X$. Suppose $W_1$ and $W_2$ are continuous, non-negative functions from $X$ into the real numbers such that, for all integers $p > 0$, $$ \int_X W_i(x)\ d\mu(x) < +\infty, \ \ i = 1,2. $$ Define the weighted $\mathcal{L}^p$-space $\mathcal{L}_W^p(X)$ as the space of measurable functions from $X$ to the reals for which the integral $$ \left(\int_X f(x)^pW(x)\ d\mu(x)\right)^{1/p} $$ is finite. In this case the function $W$ is called the weight function for $\mathcal{L}_W^p$. Furthermore, it is well known that the integral above constitutes a norm on $\mathcal{L}^p_W(X)$. Under which conditions on the weight functions $W_1$ and $W_2$ are $\mathcal{L}_{W_1}^p(X)$ and $\mathcal{L}^p_{W_2}(X)$ topologically isomorphic? (i.e. when is there a Banach space isomorphism which is also a homeomorphism with respect to the norm-topologies due to the two weight functions?)
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1$\begingroup$ By "positive definite" do you mean non negative? If the weight functions are both strictly positive and finite, then the two weighted $L_p$ spaces are isometrically isomorphic via a multiplication operator. This is easy and basic, so maybe you mean something else? $\endgroup$– Bill JohnsonCommented Jan 6, 2018 at 21:15
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$\begingroup$ @BillJohnson Yes I meant non-negative. I am very much not an expert in this area, so the basic result eluded me. For the record, what would the operator be? Using the multiplier $W_1/W_2$ (or its inverse) would lead to problems in cases where $W(x) = 0$ for some $x$. $\endgroup$– JMJCommented Jan 6, 2018 at 21:18
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$\begingroup$ Note that @BillJohnson specified that he was talking about strictly positive weight functions. $\endgroup$– LSpiceCommented Jan 6, 2018 at 21:32
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$\begingroup$ @LSpice okay. Then yes, I agree this is a trivial case. Non-negativity is the key condition. We may not assume $W(x) \neq 0$. $\endgroup$– JMJCommented Jan 7, 2018 at 0:53
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1$\begingroup$ OK - is the following sufficient for your needs? If $X$ is not very big (say, separable, metriziable, complete) - then the $L^p$ spaces associated to all purely non-atomic $\sigma$-finite measures on $X$ are isometrically isomorphic. $\endgroup$– R WCommented Jan 7, 2018 at 2:49
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The space $L^p_W(X)$ is nothing else than the $L^p$ with respect to the finite measure $\mu_W$ whose density with respect to $\mu$ is $W$. Now you can use the fact that the measure spaces determined by purely non-atomic Borel probability measures on "nice" spaces $X$ (separable, metrizable, complete) are all isomorphic to the unit interval endowed with the Lebesgue measure, which implies that for any fixed $p$ the corresponding $L^p$ spaces are all isometrically isomorphic.
