Is $\mathcal{B}(L^p,L^1)$ strictly bigger than $L^q$?

Question

Let $(\Omega, \mathscr{F}, \mu)$ be a $\sigma$-finite measure space and $1\leq q \leq \infty$. Denote by $p$ the conjugate exponent of $q$. We know, given a $g\in L^q (\Omega)$, \begin{equation} T:f \in L^p(\Omega) \mapsto gf \end{equation} defines an element of $\mathcal{B}(L^p(\Omega),L^1(\Omega))$. Moreover, by H"older's inequality, we have $\|T\|\leq \|g\|_q$. Actually, if $q<\infty$, taking $f=|g|^{q-1}\in L^p(\Omega)$, we obtain \begin{equation} \|T\|\geq \frac{\|gf\|_1}{\|f\|_p} = \frac{\|g\|_q^q}{\|g\|_q^{q/p}} = \|g\|_q, \end{equation} thus, $\|T\|=\|g\|_q$; if $q=\infty$, considering the characteristic functions $f_n=\chi_{\{|g|\geq \|g\|_\infty- \frac{1}{n}\}}$ for those sufficiently large $n$'s, we also have $\|T\|=\|g\|_\infty$. Therefore, \begin{equation} g\mapsto T \end{equation} is an isometric embedding from $L^q(\Omega)$ to $\mathcal{B}(L^p(\Omega),L^1(\Omega))$. I guess the embedding is a surjective isometry. But I have difficulty verifying or disproving the statement. Any ideas would be appreciated.

Answer 1

This is most easily seen to be wrong for $p = 1$. In that case you are essentially asking if every operator from $L^1$ to $L^1$ is a multiplication operator, which is obviously not the case unless your measure space is very special. With this in mind, you can probably already see why this in wrong also for $p > 1$. To give an example, consider the counting measure on some finite set, say with $n$ elements. Then $\mathcal{B}(L^p, L^1)$ is equal to the set of $n \times n$ matrices. The elements you describe are exactly the diagonal matrices. But obviously there are more matrices than just the diagonal ones.

Answer 2

This map can only be surjective in degenerate cases. Take $A,B\in\mathcal{F}$ with $0<\mu(A),\mu(B)<\infty$ and $\mu(A\cap B)=0$ (these may not exist, which is what I mean by "degenerate cases"). Let $E\subset L^q$ be the linear span of $1_A$. Since $E$ is finite-dimensional, there exists a closed subspace $F\subset L^p$ such that $L^p=E\oplus F$. Define $S\in B(L^p,L^1)$ in such a way that $S1_A=1_B$ and $S|_{F_q}=0$. Since $f1_A=0$ a.e. on $B$ for every $L^q$, there is no $f\in L^q$ such that $Sg=fg$ for all $g\in L^p$.