$F$ continuous and the initial value problem uniquely solvable

Question

Consider $$l^\infty = \{v=(v_n)^\infty_{n=1} :\Vert v \Vert_{l^\infty}:= \sup_n \vert v_n \vert < \infty \}.$$ Now let $p \in (0,1)$ and define the sequence $(a_n)^\infty_{n=1}$ recursively by $a_1:=1$ and $a_{n+1}:=pa_n +1$. Furthermore let $F: l^\infty \to l^\infty$ be a given mapping where $$(F(v))_1=0, \,\, \,(F(v))_{n+1}:=a_n\vert v_n\vert ^p\,\,$$

How can I show that $F$ is continous and that for $u_0 \in l^\infty$ the solution of the initial value problem $$\begin{cases} u'(t)= F(u(t)) \\ u(0)=u_0 \end{cases}$$ is unique?

Answer 1

You can show (by recurrence, for example) that $$a_k = 1 + p + \cdots + p^{k-1}$$ and thus $$\forall k \quad 1\le a_k < a_{k+1} < \frac{1}{1-p}<+\infty.$$

after the question was edited

First, we need to make sure that $F(v) \in \ell^\infty$ for $v\in\ell^\infty$. This comes from the fact that $a_n$ are bounded from above and that $|v_n|^p$ are bounded by $\|v\|^p_\infty$.

The contunuity of $F$ follows from the inequality $$\sup_n\left|(F(v)-F(w))_{n+1}\right| = \sup _na_n \left| |v_n|^p - |w_n|^p\right| \le \frac{1}{1-p} \sup _n \left| |v_n|^p - |w_n|^p\right|, $$ and the latter goes to zero as $w\to v$.