# What is the trace of this operator in $L^\infty$ (if this question make sense)?

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 !

• Suppose $f\equiv1$. Then $A_f$ is the identity over an infinite dimensional space. Its trace should be infinite. Commented May 12, 2015 at 13:06
• Without extra conditions on $f$ I think this is fairly hopeless Commented May 12, 2015 at 13:07
• math.tamu.edu/~johnson/TF3.4.pdf Commented May 12, 2015 at 13:42
• @LiviuNicolaescu Perhaps I have misunderstood, but how is that (very nice) paper relevant to the present discussion? The authors are looking at weak Hilbert spaces and $L^\infty$ is very far from being such a space: they are only looking at nuclear operators and Denis Serre's example points out that the multiplication operators considered by the OP need not even be compact Commented May 12, 2015 at 14:27
• Isn't multiplication by $f$ compact precisely if $f=0$ ? Commented May 12, 2015 at 14:32