1

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 $$f^n(t,x) \to 0 \quad\text{as $t \to 0$ uniformly in $n$}?$$

25Chars
  • 566
  • You are probably not interested in the finite dimensional case or when $f(t,\cdot)\to0$ uniformly on bounded sets, right? In your last equation, do you want $f^n(t,x)\to0$ as $t\to0$ uniformly in $n$ to hold for any $x$ or also have some uniformity wrt $x$? I'm struggling to find a counter-example in the infinite dimensional setting. – s.harp Feb 13 '17 at 19:11
  • @s.harp Yes, it's infinite-dimensional, and the convergence is only on compact sets unfortunately, and the $x$ is arbitrary. – 25Chars Feb 13 '17 at 20:07

1 Answers1

1

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.

Community
  • 1
s.harp
  • 22,576
  • Could you please say why uniform on bounded subsets would be enough for the result? – Upin Apr 12 '17 at 09:24
  • 1
    Start with some $x$ and look at the ball $B_{2|x|}(0)$. Now for any $\epsilon$ you've got a $T$ so that if $t>T$ the image of $B_{2|x|}(0)$ under $f(t,\cdot)$ lies in $B_\epsilon(0)$. Wlog assume that $|x|>\epsilon$ so $B_\epsilon(0)\subset B_{2|x|}$ and then iteratively $f^n(t,x)$ lies in $B_\epsilon(0)$ for all $n$ whenever $t>T$. – s.harp Apr 12 '17 at 10:51