In this paper of Zubelevich the author consider a a scale of Banach spaces $\{(E_{s},\|\cdot\|_{s}):0<s<1\}$, 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 $0<s<1$. Then, fixed $a>1$, defines the (open) triangle $$ \Delta :=\{(\tau,s)\in\mathbb{R}^{2}:\tau>0,0<s<1,1-s-a\tau>0 \}, $$ 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 $B_{\tau,s}:=\{u\in E: \|u\|_{\tau,s}<r\}$, 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?

Thank you very much in advance for your comments.

  • $\begingroup$ 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 $\tau_1\ne\tau_2$? $\endgroup$ – daw Jan 10 at 7:32
  • $\begingroup$ 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... $\endgroup$ – user123043 Jan 10 at 8:18
  • 1
    $\begingroup$ How do you define the intersection of function spaces, where the functions have different domain? I.e, what is $C([0,1];\mathbb R) \cap C([0,2],\mathbb R)$. $\endgroup$ – daw Jan 10 at 11:15
  • $\begingroup$ Yes, I agree with you, How it can be defined that intersection? I am reading the cited paper, but I feel that the definition of E can has "some hole", no? Thanks @daw $\endgroup$ – user123043 Jan 10 at 11:28
  • $\begingroup$ It seems I met the author at Russian scientific forum where he is very active and writes different things. So I think there are big chances that he will answer your letter. $\endgroup$ – Alex Ravsky Jan 10 at 15:45

Your Answer

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

Browse other questions tagged or ask your own question.