Let $\Omega$ be a closed domain in $\mathbb{R}^N$, and $\lambda$ the corresponding Lebesgue measure. We define in $(L^\infty(\Omega), \Vert \cdot \Vert_\infty)$ the following operator :

\begin{array}{rccc} A_f : &L^\infty(\Omega)&\longrightarrow & L^\infty(\Omega) \\ &g & \longmapsto & f g \end{array}

where $f \in L^\infty(\Omega)$ is fixed.

This operator is well-defined, and is continuous for the $\Vert \cdot \Vert_\infty$ topology, with $||| A_f |||=\Vert f \Vert_\infty$.

I have the two following questions :

1) Can we define a trace for $A_f$? I know that in Hilbert space we can define the trace for the operators lying in the trace class, but here we face an operator in a non-relexive non-separable Banach space. So I have no idea of if it is a well-posed question.

2) If the trace of $A_f$ does have a sense, what is it? I feel that we should have $tr(A_f)=\int_\Omega f(x) \ dx$.

What do you think? This operator looks really simple, so if the trace can be defined in arbitrary Banach spaces for a class of nice operators, I have some hope that $A_f$ lives therein.

In the particular case of $\Omega$ being bounded (i.e. of finite measure), by using the inclusion $L^\infty(\Omega) \subset L^2(\Omega)$, we see that $A_f$ is just the restriction on $L^\infty(\Omega)$ of the continuous operator $g \mapsto \langle f,g \rangle$ induced by the usual scalar product of $L^2(\Omega)$. So maybe in that case we could pass through the trace theory in that Hilbert space, but I am not sure that it leads to the 'same' trace since the norm topologies of $L^\infty(\Omega)$ and $L^2(\Omega)$ doesn't coincide on $L^\infty(\Omega)$ (just look at the separability) ...

Thank you in advance !

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