# When multiplication operator is an bounded below/ open mapping/isometry/quotient map?

Let $(\Omega,\Sigma)$ be a $\sigma$-finite measurable space. Let $\mu,\nu\in \mathcal{M}(\Omega)$ be $\sigma$-additive measures on $\Omega$. Assume we are given $p,q\in[1,+\infty]$ and a measurable fnction $g:\Omega\to\mathbb{C}$. Which conditions on $p,q,g,\mu,\nu$ are necessary and sufficient for multiplication operator $$T:L_p(X,\mu)\to L_q(X,\nu): f\mapsto g\cdot f$$ to be $\langle$ bounded below/open mapping/isometry/quotient map$\rangle$ ?

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Here is my article where the question is fully answered.

The main result is the following: every $\langle$ bounded below / isometric $\rangle$ operator is invertible from the left and every $\langle$ open mapping / quotient mapping $\rangle$ is invertible from the right.

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Except the case when the space consist of one atom, 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., $p=q$. Note that this argument still works for infinite $p$ or $q$.