3
$\begingroup$

Fix a non-empty open domain $\Omega\subseteq \mathbb{R}^d$ with compact closure, and a finite Borel measure $\mu$ on its closure $\overline{\Omega}$.

In Halmos' book it is shown that:

Classical Result: *For any bounded function $f\in L^p_{\mu}(\Omega;\mathbb{R})$ and every $\epsilon >0$, there exists a continuous function $g$ such that $$ \int_{x \in \Omega} |f(x)-g(x)|^p \mu(dx) < \epsilon. $$*

Reference/Question: Is there an analogue of this result for the Musielak–Orlicz spaces?

More specifically, I'm wondering if $p:\Omega\rightarrow \mathbb{R}$ is a measurable function satisfying the usual conditions (for example see this paper), and $ f \in L^{p(x)}_{\mu}(\Omega),$ then for every $\epsilon>0$, can we find a continuous function $g$ on $\overline{\Omega}$ such that $$ \|f-g\|_{\mu,p}<\epsilon, $$

Background: Where in the above: $$ L^{p(x)}_{\mu}(\Omega)\triangleq \left\{ f:\Omega \rightarrow \mathbb{R}: \mbox{f is measurable and } \int_{x \in \overline{\Omega}} |f(x)|^{p(x)} \mu(dx)<\infty \right\}, $$ and can be seen to be a Musielak–Orlicz space space under the Luxemburg norm $\|\cdot\|_{\mu,p}$ defined by $$ \|f\|_{\mu,p}\triangleq \inf\left\{ \lambda >0 : \int_{x \in \overline{\Omega}} \left(\frac{|f(x)|}{\lambda}\right)^{p(x)} \mu(dx) \leq 1 \right\} . $$

$\endgroup$

1 Answer 1

3
$\begingroup$

Yes this is true (in the variable exponent's case which was the objective of the question).

See Theorem 3.4.12 of this book: "Lebesgue and Sobolev spaces with variable exponents" by "Lars Diening, Petteri Harjulehto,Peter Hästö, and Michael Růžička".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.