Uniform convergence of iterated function Let $f:\mathbb{R} \times X \to X$ be a mapping where $X$ is separable Banach spaces. We know that $f(t,\cdot)$ is continuous for fixed $t$ and also
$$f(t,x) \to 0 \quad\text{as $t \to 0$ uniformly in $x$ on compact subsets $K \subset X$}.$$
Define $f^n(t,x) = f(t, \cdot)\circ f(t,\cdot) \circ ... \circ f(t,x)$, which is composition in the second argument $n$ times.  
Does $f^n$ satisfy the property 
$$f^n(t,x) \to 0 \quad\text{as $t \to 0$ uniformly in $n$}?$$
 A: Consider a sequence of bounded linear operators in $\ell^1(\Bbb N)$
$$f_k(x)=2P_{k}+2^{-k}\tau_{k,1}$$
here $P_k$ is the usual projection onto $\mathrm{span}\{e_k\}$ and $\tau_{i,j}$ is the linear continuation of
$$\tau_{i,j}(e_l)=\begin{cases} e_i & l=j\\0&\text{otherwise}\end{cases}$$
Let $K$ be a compact subset:
$$
\sup_{x\in K}\|f_k(x)\|=\sup_{x\in K}(|2^{-k}x_1+2x_k|)≤\sup_{x\in K}\left(2^{-k}|x_1|+2|x_k|\right)
$$
This converges to zero (see here). So we have that $f_k$ converges uniformly to zero on compacta. On the other hand
$$f_k(e_1)=2^{-k}e_k\qquad f_k^2(e_1)=2^{-k+1}e_k\quad...\quad f^n_k(e_1)=2^{-k+n}e_k$$
So $\sup_{n\in\Bbb N}\|f_k^n(e_1)\|=\infty$ does not converge to zero.
Connect this to your formulation by having $f(t,x)=f_{\lfloor 1/t\rfloor}(x)$.
If you consider $f(t,\cdot)$ to converge uniformly to zero on bounded subsets of $X$ then the conclusion holds. In the finite dimensional case uniformly on compacta and uniformly on bounded sets is the same, so in the finite dimensional case you have both.
