Fix $1<p<\infty$. Let us define the increasing sequence $(a_n)_{n=1}^\infty$ by the rules $a_1=1$ and $a_{n+1}=a_n+n^{-p}$ (so that $a_{n+1}=\sum_{j=1}^nj^{-p}$). Set $$r=\lim_{n\to\infty}a_n=\sum_{j=1}^\infty\frac{1}{j^p}.$$ We define the Cesaro function space $Ces_p[0,r]$ as the Banach space of all measurable functions $f:[0,r]\to\mathbb{R}$ satisfying $$\|f\|_{Ces_p[0,r]}=\left[\int\limits_0^r\left(\frac{1}{x}\int\limits_0^x|f(t)|\;dt\right)^p\;dx\right]^{1/p}<\infty$$ We also define the Cesaro sequence space $ces_p$ as the Banach space of all scalar sequences $(c_n)_{n=1}^\infty\in\mathbb{R}^\mathbb{N}$ satisfying $$\|(c_n)_{n=1}^\infty\|_{ces_p}=\left[\sum_{k=1}^\infty\left(\frac{1}{k}\sum_{n=1}^k|c_n|\right)^p\right]^{1/p}<\infty$$
For $n\in\mathbb{N}$, let $\chi_n$ denote the indicator/characteristic function $$\chi_n(t)=\left\{\begin{array}{ll}1,&\text{ if }t\in[a_n,a_{n+1})\\0,&\text{ otherwise.}\end{array}\right.$$ Now set $$f_n=\frac{a_n}{a_{n+1}-a_n}\cdot\chi_n,\;\;\;n\in\mathbb{N}.$$ Using elementary techniques, I have been able to prove the following.
Theorem 1. There is $\delta\in(0,1)$ with $$\delta\|(c_n)_{n=1}^\infty\|_{ces_p} \leq\|\sum_{n=1}^\infty c_nf_n\|_{Ces_p[0,r]} \leq\|(c_n)_{n=1}^\infty\|_{ces_p}\;\;\;\text{ for all }(c_n)_{n=1}^\infty\in c_{00}$$ (where $c_{00}$ denotes the space of scalar sequences with finitely many nonzero entries).
Consequently, the Cesaro function space $Ces_p[0,r]$ contains an isomorphic copy of the Cesaro sequence space $ces_p$.
Question 1. Is the above isomorphic copy of $ces_p$ complemented in $Ces_p[0,r]$?
My idea is to define a projection $P$ by the rule $$(Pg)(t)=\sum_{n=1}^\infty n^p\int\limits_{a_n}^{a_{n+1}}g(s)\;ds\cdot\chi_n(t).$$ If $P$ is bounded on $Ces_p[0,r]$, then it is a projection onto $[f_n]_{n=1}^\infty$.
Question 2. Is the above operator $P$ bounded on $Ces_p[0,r]$?
Thanks guys!