Consider two metric spaces $X$ and $Y$ and a sequence of functions $f_n\colon X\to Y$ together with a function $f\colon X\to Y$. Assume, all $f_n$ and $f$ have inverse functions $g_n$ and $g$, say. It is true, that if $g$ is continuous and $f_n\to f$ compactly, i.e. uniformly on every compact set, then also the inverse functions $g_n$ converge to $g$ compactly.

In Uniform convergence of functions, Spring 2002 they gave a nice proof for a similar result, using uniform convergence of $f_n$ and uniform continuity of $f$ to conclude that $g_n$ converge uniformly. The idea was to show that $f\circ g_n$ converges to $f\circ g$ uniformly and then conclude that also $g_n = g\circ f\circ g_n$ converge uniformly to $g = g\circ f\circ g$.

I would be interested if it is possible to adapt this prove to compact convergence. The main problem I see is the following: Showing $f\circ g_n$ converges to $f\circ g$ compactly is equivalent to showing $f\circ g_n$ converges to $f_n\circ g_n$ compactly. But for fixed compact set $K\subset Y$ I do not see how we can guarantee that all $g_n(K)$ stay in the same compact set $L\subset X$. Hence, we can not apply compact convergence of $f_n$. Are there any ideas for finding the compact set $L$?

  • $\begingroup$ No continuity assumptions? $\endgroup$ – Daniel Fischer Nov 13 '15 at 10:27
  • $\begingroup$ No, I think it already works if we only assume $g$ to be continuous (or uniformly continuous for uniform convergence). $\endgroup$ – Marcel Nov 13 '15 at 10:31
  • $\begingroup$ I want to elaborate on the importance of continuity a bit. For simplicity, I do so in case of uniform convergence. To show $f\circ g_n$ converges to $f\circ g$ uniformly, it suffices to show $f_n\circ g_n$ converges uniformly to $f\circ g_n$ (by triangle inequality). The last claim follows from uniform convergence of $f_n$. To show the second claim indeed just uniform convergence of $f\circ g_n$ to $f\circ g$ and uniform continuity of $g$ is needed. So no more continuity assumptions are necessary. $\endgroup$ – Marcel Nov 13 '15 at 13:55
  • $\begingroup$ @Marcel do you have a reference for a different proof of this statement? Thanks! $\endgroup$ – Aru Ray Jan 9 at 14:27
  • $\begingroup$ @AruRay No, I don't. Sorry. $\endgroup$ – Marcel Feb 15 at 9:35

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