Let $(X,d)$ be a proper metric space (closed balls are compact), and $r :[0, \infty) \rightarrow X$ be a continuous and proper (inverse image of compact is compact) map, with $r(0) = x_0$. Therefore, given a radius $R$, there exists $N_R > 0$ such that $r([N_R, \infty))$ avoids the closed ball $\overline{B}(x_0 ; R)$. Can the choice $R \mapsto N_R$ be made continuous in $R$? My last question came out of a very naive (and unsuccessful) attempt of doing so.

  • $\begingroup$ How is that choice made for large r when X is bounded? $\endgroup$ – William Elliot Apr 12 '18 at 3:53
  • $\begingroup$ @WilliamElliot $X$ is non-compact because the map $r$ is assumed to be proper. On the other hand, the closed balls are assumed to be compact. $\endgroup$ – Cihan Apr 12 '18 at 4:51
  • $\begingroup$ What if r = sin? $\endgroup$ – William Elliot Apr 12 '18 at 6:36
  • $\begingroup$ Consider X = {0}. $\endgroup$ – William Elliot Apr 12 '18 at 6:38
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    $\begingroup$ @DanielWainfleet You mean $\inf$ instead of $\sup$. Still showing that this function is continuous does not seem to be trivial. Michael selection theorem might be helpful here. The multivalued function $f(R)=\{x\ |\ r([x,\infty))\cap D(r(0), R)=\emptyset\}$ may work. $X$ is paracompact, values are nonempty and convex subsets of $\mathbb{R}$. The only problematic property is lower hemicontinuity. $\endgroup$ – freakish Apr 12 '18 at 9:59

Yes. Start with the function $f:(0, \infty) \to (0,\infty)$ defined by $$ f(R) = 1 + \inf\{t : r([t, \infty))\cap \overline{B}(x_0, R) = \varnothing \} $$ This function may be discontinuous, but it is nondecreasing. For any such function there exists a continuous function $g:(0, \infty) \to (0,\infty)$ such that $g(R)\ge f(R)$ for all $R$. (One may say that $f$ has a continuous majorant.)

Proof: First put a staircase on top of $f$, for example $h(x) = f(\lceil n\rceil)$ does the job (it's a piecewise constant function). Then roll a carpet over the staircase, producing a piecewise linear function majorizing $h$. Putting this together in a formula, $$ g(n+\alpha) = (1-\alpha)f(n+1) +\alpha f(n+2),\quad n\in\mathbb Z, \quad \alpha\in [0, 1) $$ The fact that $g\ge f$ follows from $$ (1-\alpha) f(n+1) + \alpha f(n+2) \ge (1-\alpha) f(n+\alpha) + \alpha f(n+\alpha) = f(n+\alpha)$$ The continuity follows from $g$ being linear on each segment $[n, n+1)$ and being continuous at the endpoints: $g(n) = f(n+1)$ for all integers $n$.

  • $\begingroup$ The definition of $g$ says $g(n) = f(n+2)$. Did you mean to interchange $\alpha$ with $1-\alpha$? Neat solution btw with a generally useful technique: dominating any increasing function with a continuous one. $\endgroup$ – Cihan Apr 13 '18 at 3:31
  • $\begingroup$ Yes, I had the coefficients misplaced. Fixed now. $\endgroup$ – user357151 Apr 13 '18 at 3:43

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