The setting : let $(\Omega,\mu)$ $\sigma$-finite measure space and let $M_\phi : L^2(\Omega,\mu) \to L^2(\Omega,\mu)$ the multiplication operator with $\phi \in L^{\infty}(\Omega,\mu)$

I want to show :

If $M_{\phi_{i}} \to M_\phi $ in weak operator topology, then $\phi_i \to \phi$ in weak*-topology

I already managed to show the reverse statement.

I don't know if this helps or even is true : Maybe I can write every $f \in L^1$ as product of two functions in $L^2$ ?

  • $\begingroup$ Which weak$^*$ topology do you mean? The one which comes from $L^1$? $\endgroup$ – tomasz Feb 6 '17 at 23:34
  • $\begingroup$ The one coming from $L¹$ $\endgroup$ – Chen Huang Feb 6 '17 at 23:35

As alluded to in your question, the hardest part is writing an $L^1$ function as a product of two $L^2$ functions. But this turns out to be easier than expected.

Suppose $M_{\phi_i}$ is WOT-convergent to $M_\phi$, and let $f\in L^1$ be given. Then we can write $f=|f|e^{i\theta}$, where $\theta$ is a measurable function. Now define \begin{align*} g&=|f|^{1/2}e^{i\theta}, \\ h&=|f|^{1/2}. \end{align*} Then $g,h\in L^2$ and we have $$\langle M_{\phi_i}g,h\rangle=\int\phi_i|f|e^{i\theta}\ d\mu =\int\phi_if\ d\mu. $$ By hypothesis, $\langle M_{\phi_i}g,h\rangle\to\langle M_\phi g,h\rangle$, and thus $$ \int\phi_if\ d\mu\to\int\phi f\ d\mu. $$ Since $f\in L^1$ was arbitrary, we know $\{\phi_i\}$ is weak$^*$-convergent to $\phi$.


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