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The question I have in mind is (see here, page 60, the solution is at page 297):

Assume $f_{n}$ is a sequence of functions from a metric space $X$ to $Y$. Suppose $f_{n}\rightarrow f$ uniformly and has inverse $g_{n}$. Now assume $f$'s inverse $g$ is uniformly continuous on $Y$. Prove that $g_{n}\rightarrow g$ uniformly.

I could not prove it using standard techniques as I do not know how to bound $|g_{n}(y)-g(y)|$ when $n$ becomes very large. The authors argue that the convergence of $g_{n}(y)\rightarrow g(y)$ is similar to $f(g_{n}(y))\rightarrow f(g(y))$ because the mapping by a uniformly convergent function series keeps uniform convergence. Thus they give the following argument that $$d(f(g(y)),f(g_{n}(y)))=d(y,f(g_{n}(y)))\le d(y,f_{n}(g_{n}(y)))+d(f_{n}(g_{n}(y)),f(g_{n}(y)))=d(f_{n}(g_{n}(y)),f(g_{n}(y)))$$

So since $f_{n}\rightarrow f$ uniformly by hypothesis the statement is proved. My question is: Is the step of substituting $|g_{n}(y)-g(y)|$ by $|f(g(y))-f(g_{n}(y))|$ really justified? I could not get the "keep uniform convergence" thing the author is talking about. But I also could not come up with a better proof.

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If $h_n\colon S\to S'$ converges uniformly to $h$ on $S$ and $\varphi\colon S'\to S''$ is uniformly continuous on $S$, fix $\varepsilon>0$. There is a $\delta>0$ such that if $d_{S'}(x,y)\leq \delta$ then $d_{S''}(\varphi(x),\varphi(y))\leq\varepsilon)$. Now, we use the fact that there is an integer $n_0$ such that for all $n\geq n_0$, we have $$\sup_{x\in S}d_S(h_n(x),h(x))\leq \delta.$$ Then $$\sup_{x\in S}d_S(\varphi(h_n(x)),\varphi(h(x)))\leq\varepsilon.$$ Now, we apply this result with $\varphi=g$ and $h_n:=f\circ g_n$.

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In translation we have $|f(g_{n})-f(g)|\le \delta$ when $n\ge N$, and $|g(x)-g(y)|\le \epsilon$ when $|x-y|\le \delta$. Thus we have $|g(y)-g_{n}(y)|=|g(f(g(y))-g(f(g_{n}(y))|\le \delta$ once $n\ge N$. But this feels wierd - why would $f(g_{n})$ converges uniformly to $g$? –  Bombyx mori Jul 26 '12 at 9:10
    
Sorry I was confused. I should have written $f(g_{n})$ to $f(g)=I$. –  Bombyx mori Jul 26 '12 at 9:31
    
Davide, where are you..? –  Bombyx mori Jul 26 '12 at 9:39
    
It has been shown that $f\circ g_n$ converges uniformly to $f\circ g$. By what I showed, $g\circ f\circ g_n$ converges uniformly to $g\circ f\circ g$, that is $g_n$ converges uniformly to $g$. –  Davide Giraudo Jul 26 '12 at 9:45
    
Davide, I do not think it is given $f\circ g_{n}$ converges to $I$ uniformly. We only know $f_{n}\rightarrow f$ uniformly and $g_{n}\rightarrow g$, the later's uniformity is skeptical unless we proved it. –  Bombyx mori Jul 26 '12 at 9:47
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