0
$\begingroup$

We consider the set $\mathcal{PC}([-r,0],X)$

$$\mathcal{PC}([-r,0],X):=\{\varphi:[-r, 0] \rightarrow X: \varphi \text{ is continuous everywhere except for a finite number of points } t_* \text{ such that } \varphi\left(t_*^{-}\right) \text{ and } \varphi\left(t_*^{+}\right) \text{ exist and } \varphi\left(t_*\right)=\varphi\left(t_*^{-}\right) \}$$

equipped with the following norm $$\|\varphi\|=\displaystyle\int_{-r}^{0}\|\varphi(\theta)\|d\theta. $$ We consider a space $\mathcal{B}_{g}$ as $$ \mathcal{B}_{g}:=\left\{\varphi:(-\infty, 0] \rightarrow X:\left.\varphi\right|_{[-r, 0]} \in \mathcal{PC}([-r,0],X) \text { and } \int_{-\infty}^0 g(\theta)\|\varphi(\theta)\| d \theta<+\infty\right\}. $$ Then, we set $$ \|\varphi\|_{\mathcal{B}_{g}}=\int_{-\infty}^0 g(\theta)\|\varphi(\theta)\| d \theta, \text { for all } \varphi \in \mathcal{B}_{g}. $$ where $g:(-\infty,0]\longrightarrow (0,+\infty)$ is a continuous function such that $\rho=\displaystyle\int_{-\infty}^{0}g(s)ds<+\infty.$ I made every effort to demonstrate the completeness of the space $\mathcal{B}_{g}$. However, I encountered a significant challenge when attempting to prove that the restriction of the limit of the Cauchy sequence, taken in $\mathcal{B}_{g}$, onto the interval $[-r,0]$, belongs to $\mathcal{PC}([-r,0],X)$. I would greatly appreciate any assistance or guidance, please.

Improve this question
$\endgroup$
4
  • $\begingroup$ I find it hard to believe that $\mathcal{B}_g$ is complete. The inverse of the Cantor function (take a caglad version) can be expressed as the uniform limit of caglad functions , each of which with finitely many discontinuities. Under you hypotheses, uniform convergence implies $\mathcal{B}_g$ convergence, but the inverse Cantor function is not in $\mathcal{B}_g$ as you defined it, as it has infinitely many discontinuity points. $\endgroup$
    – Willie Wong
    Commented Jun 28, 2023 at 20:17
  • $\begingroup$ Thank you for your response. Could you please provide more explicit details or clarification? $\endgroup$
    – Mathlover
    Commented Jun 28, 2023 at 20:33
  • $\begingroup$ Of course this space can't be complete . Think of $X=\mathbb R$, $g:=e^\theta$ (if you want an example with $\rho<\infty$, or just $g:=1$ to make it simpler). Then $B_g$ is dense in $L^1$, where you can find functions that are everywhere discontinuous and unbounded. $\endgroup$
    – Pietro Majer
    Commented Jun 30, 2023 at 17:04
  • $\begingroup$ Thank you sir for your help. $\endgroup$
    – Mathlover
    Commented Jul 2, 2023 at 12:25

0

Reset to default

You must log in to answer this question.