# Is this seminormed space (intersection of Banach spaces) non-empty?

In this paper of Zubelevich the author consider a a scale of Banach spaces $$\{(E_{s},\|\cdot\|_{s}):0, that is to say, each $$(E_{s},\|\cdot\|_{s})$$ is a Banach space and

$$E_{s+\delta}\subseteq E_{s},\quad \|\cdot\|_{s}\leq \|\cdot\|_{s+\delta}, \quad s+\delta<1, \delta>0,$$ for each $$0. Then, fixed $$a>1$$, defines the (open) triangle $$\Delta :=\{(\tau,s)\in\mathbb{R}^{2}:\tau>0,00 \},$$ and the seminormed space $$E:=\bigcap_{(s,\tau)\in \Delta} C([0,\tau],E_{s}),$$ endowed the family of norms $$\|u\|_{\tau,s}:=\max_{0\leq t\leq \tau}\|u(t)\|_{s}$$. As usual, $$C([0,\tau],E_{s})$$ denotes the space of the continuous maps $$u:[0,\tau]\longrightarrow (E_{s},\|\cdot\|_{s})$$. So, $$E$$ is a (locally convex) topological space with a basis of the topology given by the open balls $$B_{\tau,s}:=\{u\in E: \|u\|_{\tau,s}, for each $$r>0$$.

My question is the following: Is $$E$$ non-empty? Or, more precisely, under that conditions $$E$$ is not a "trivial" (for instance, a single point) space?

• Aren't constant functions that have values in $E_1$ contained in $E$? Also, what is the intersection of $C([0,\tau_1],E_{s_1}$ with $C([0,\tau_2],E_{s_2}$ for $\tau_1\ne\tau_2$? – daw Jan 10 at 7:32
• Thanks @daw. $E_{1}$ "a priori" could not be a Banach space, not? i.e, may not be "well defined" in the escale. I do not undestand at all your second question... – user123043 Jan 10 at 8:18
• How do you define the intersection of function spaces, where the functions have different domain? I.e, what is $C([0,1];\mathbb R) \cap C([0,2],\mathbb R)$. – daw Jan 10 at 11:15