6
$\begingroup$

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 $L^2$ ?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\phi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.