When multiplication operator is an isometry/ bounded below?

Question

Let $X$ be a locally compact topological space. Let $\mu,\nu\in M(X)$ - regular finite $\sigma$-additive Borel measures and $\nu$ is absolutely continuous with respect to $\mu$. Let $p,q\in[1,+\infty]$ and $g:X\to\mathbb{C}$ a measurable function such that multiplication operator $$ T:L_p(X,\mu)\to L_q(X,\nu): f\mapsto g\cdot f $$ is well defined.

Question №1. Does there exist a criterion for operator $T$ to be isometric? This question is almost answered by Lukas Geyer. It is remains to consider case when $p$ or $q$ or both are equal to $\infty$, and this is quite easy. The necessary condition is $p=q$. And for $p=q$ description of isometries is given in this article.

Question №2. Does there exist a criterion for operator $T$ to be bounded from below? For the case $p=q$ the necessary and sufficient conditions are

• $d\nu/d\mu>0$ almost everywhere on $X$

• zero is not in the essential range of $g(d\nu/d\mu)^{-1/p}$

The remaining case is $p\neq q$.

Thank you for taking time.

Answer 1

Except for trivial cases this operator is never an isometry if $p \ne q$. If $A$ and $B$ are disjoint sets with $\mu(A), \mu(B) \in (0,\infty)$ (the "trivial case" refers to the case if no such sets exists), then there exist functions $f_A$ supported on $A$ and $f_B$ supported on $B$ with $$ \int |f_A|^p \, d\mu = \int |f_B|^p \, d\mu = 1. $$ Then $$ \int |f_A+f_B|^p \, d\mu = 2, $$ so the $p$-norms with respect to $\mu$ of $f_A$, $f_B$, and $f_A+f_B$ are $1$, $1$, and $2^{1/p}$, respectively.

If $T$ is an isometry, then the $q$-norms with respect to $\nu$ of $T(f_A)$, $T(f_B)$ and $T(f_A+f_B)$ also have to be $1$, $1$, and $2^{1/p}$. This implies $$2^{q/p} = \int |g\cdot(f_A+f_B)|^q \, d\nu = \int|g\cdot f_A|^q \, d\nu + \int|g\cdot f_B|^q\, d\nu = 1+1 = 2,$$ so that $q/p=1$, i.e., $p=q$.