I'm trying to understand the demonstration above.
For 1, if $g\in L^{2}(\mathbb{R}^n)$ and $g\notin L^p(\mathbb{R}^n)$, it is clear that \begin{align} \|f\|_{L^2} = \|e^{-it_{0}\Delta}g\|_{L^2} &= \|g\|_{L^2} < +\infty\\ \|e^{it_0\Delta}f\|_{L^p} = \|e^{it_0\Delta}e^{-it_{0}\Delta}g\|_{L^p} &= \|g\|_{L^p} = +\infty, \end{align} Moreover, I don't know if I should choose a specific function that satisfies those hypotheses. For instance:
If $g(x) = \frac{1}{|x|^{\frac{1}{2}-\varepsilon}}\chi_{[0,1]}$, then it belongs $L^2(\mathbb{R})$, but not to $L^p(\mathbb{R})$ for $p>2$. I don't know how to generalise this to $\mathbb{R}^n$.
For 2, I don't understand how he concludes that $e^{it\Delta}\notin H^{s'}(\mathbb{R}^n)$.
Ad 1: I suppose you could just take $$ g(x) = \vert x \vert^{\varepsilon - \frac{1}{2}} \mathbf{1}_{B_1(0)}.$$ Ad 2: The computation in the proof shows that if $e^{it\Delta}f\in H^{s_0}(\mathbb{R}^n)$, then also $f = e^{-it\Delta}(e^{it\Delta})f\in H^{s_0}(\mathbb{R}^n)$. Now, if you suppose that $e^{it\Delta}f\in H^{s'}(\mathbb{R}^n)$, then ultimately $f\in H^{s'}(\mathbb{R}^n)$, which is a contradiction to the assumption that $f\not\in H^{s'}(\mathbb{R}^n)$.