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 $E^D$.


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?

  • 2
    $\begingroup$ Just to clarify: are you asking for a linear homeomorphism between these $L^p$-spaces, or merely a homeomorphism between the underlying topological spaces? $\endgroup$ – Yemon Choi Jun 21 '19 at 21:40
  • 4
    $\begingroup$ 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. $\endgroup$ – Piotr Hajlasz Jun 22 '19 at 5:56
  • 1
    $\begingroup$ 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 $p$? $\endgroup$ – Gerald Edgar Jun 22 '19 at 14:21
  • $\begingroup$ My comment has $d_Y$ a bounded metric and $\mu$ a finite measure. Think about that case. $\endgroup$ – Gerald Edgar Jun 23 '19 at 12:10
  • 1
    $\begingroup$ @AlexRavsky Ofcourse, I put in those important details. $\endgroup$ – BLBA Mar 6 '20 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.