I encountered the following space as a natural space for setting up a certain problem: $$ S_m^p = \{f \colon I \to \mathbb{R} \text{ measurable }; m^{f} \in L^p(I)\} $$ Here, $I$ is an open bounded interval and $m$ is a positive real number. The notation $S_m^p$ is ad hoc.

Is there any other characterization of the space $S_m^p$? What about of $$ S^p = \bigcap_{m \in \mathbb{R}_+} S^p_m? $$

Some remarks

  • $S_1^p$ is quite large and not very interesting.
  • $S_m^p$ is not generally a vector space: $\alpha f \in S_m^p$ if and only if $f \in S^p_{m^\alpha}$, and $f+g \in S_m^p$ if and only if $m^f m^g \in L^p$, and $L^p$ is not generally closed with respect to multiplication.
  • Clearly $L^\infty \subseteq S^\infty$. For the other direction, consider any $m_+ > 1$ and $m_- < 1$. These give an upper and a lower bound for $f$, respectively. Hence $L^\infty = S^\infty$.
  • Presumably for $1 \leq m_1 \leq m_2$ we have $S^p_{m_2} \subseteq S^p_{m_1}$ (and for $m_2 \leq m_1 \leq 1$), and for $p_1 \leq p_2$ we have $S^{p_2}_{m} \subseteq S^{p_1}_{m}$.
  • 3
    $\begingroup$ To get linear spaces, see the Orlicz space material. https://en.wikipedia.org/wiki/Birnbaum–Orlicz_space Since the map $x \mapsto m^x$ fails the $\Delta_2$ condition, to get an actual linear space you have to be more clever than $\{f \colon I \to \mathbb{R} \text{ measurable }; m^{f} \in L^p(I)\}$ $\endgroup$ – Gerald Edgar Apr 10 '18 at 12:55

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