I need some help with this question:
We consider the dilatation operator:
$T: \mathbb{R^{+}}\to \mathcal{L}(L^p(\mathbb{R}),L^p(\mathbb{R}))$
$\;\;\;\;\;\;\delta\to D_{\delta}f=\frac{1}{\delta}f(\frac{x}{\delta})$
I have to study its measurability. I have to see if it is SOT-measurable, but not $||\cdot||$-measurable.
Our definitions are:
$X=\cal{L}$($E_1$,$E_2$)
$f:\Omega\to X$$\;,\;\;$$(X,\Sigma,\mu)$ measure space.
$\bullet\; ||\cdot||_{X}$-measurable: $\exists s_n:\Omega\to X$, simple, and $\exists A\in\Sigma,\;\mu(A)=0$, with $||s_n(w)-f(w)||_{X}\xrightarrow[n\to\infty]{}0\;\;\forall w\notin A.$
$\bullet\;$ SOT-measurable: $\forall e_1\in E_1$, the function $w\to f(w)(e_1)\;\;$ ($\Omega\to E_2$) is $||\cdot||_{E_2}$-measurable.
Thanks a lot in advance for any help.