# Homeomorphism between $L^p$-spaces on metric spaces and $L^p$-spaces on Euclidean space

Setup:

Fix $$p \in [1,\infty)$$.
Let $$(X,d_X,x_0)$$ and $$(Y,d_Y,y_0)$$ be complete pointed metric spaces and $$\mu$$ be Borel. Let $$E^n,E^D$$ be Euclidean spaces of respetive dimensions $$n$$ and $$D$$ and suppose that $$f_1:(X,d_X)\rightarrow E^D$$ and $$f_2:E^n\rightarrow (Y,d_Y)$$ are homeomorphisms satisfying $$f_1(x_0)=0 \mbox{ and } f_2(0)=y_0 .$$

Let $$\mu$$ be a $$\sigma$$-finite Borel measure on $$(X,d_X)$$ Define the $$L^p$$-like space $$L^p_{loc}(X,Y;\mu)$$ as being the set of all measurable functions $$f$$ for which $$\int_{x \in K} d^p_Y(f(x),f(x_0)) \mu(dx)<\infty,$$ for every compact subset $$K\subseteq X$$.

Suppose that $$X$$ admits a countable cover of compact sets $$\{K_n\}_{n \in \mathbb{N}}$$. Then this can be made into a complete metric space (standard result) when endowed with the metric $$d(f,g)= \sum_{n=1}^{\infty} \frac1{2^n}\sqrt[p]{\int_{x \in K_n} d^p_Y(f(x),g(x)) \mu(dx)} .$$ Let $$L^p(E^D;E^n;\nu)$$ denote the Bochner-space of all $$\nu$$-measurable functions from $$E^D$$ to $$E^n$$. Denote by $$f^{\star}(\mu)$$ the push-forward measure on $$E^D$$.

Question:

1) Is the map $$F$$, defined by: \begin{aligned} F: L^p_{loc}(X,Y;\mu)&\rightarrow L^p_{loc}(E^D;E^n;f^{\star}(\mu))\\ f &\mapsto f_2\circ f\circ f_1, \end{aligned} a homeomoprhism?

2) Are the continuous functions dense in this space?

• Just to clarify: are you asking for a linear homeomorphism between these $L^p$-spaces, or merely a homeomorphism between the underlying topological spaces? – Yemon Choi Jun 21 '19 at 21:40
• Just for the record: all separable Banach spaces are homeomorphic. Proving that your space is homeomorphic to $L^p$ does not seem to be very useful if you do not know anything about the homeomorphism. – Piotr Hajlasz Jun 22 '19 at 5:56
• As an example, suppose $X$ is $[0,1]$ with he usual metric and Lebesgue measure. Suppose $d_Y$ is a bounded metric. (I assume $d_Y^p$ should be inside the integrals.) But then $L^p(X,Y;\mu)$ is all measurable functions. And I guess $d(f,g)$ is a metric for convergence in measure. So perhaps in that case your space does not depend on $p$? – Gerald Edgar Jun 22 '19 at 14:21
• My comment has $d_Y$ a bounded metric and $\mu$ a finite measure. Think about that case. – Gerald Edgar Jun 23 '19 at 12:10
• @AlexRavsky Ofcourse, I put in those important details. – BLBA Mar 6 '20 at 11:52