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I have a question: Consider two sequences of continuous, bijective functions $f_n$ and $g_n$ mapping $\mathbb{R}\to\mathbb{R}$. I know that for every compact $K\in\mathbb{R}$ that

$$\lim_{n\to\infty}\int_K |f_n(x)-g_n(x)| \mathrm d x =0.$$

Furthermore the limit $f$ is surjective and continuous and

$$\lim_{n\to\infty}\int_K |f_n(x)-f(x)| \mathrm d x =0,$$

Now, does this also hold for the inverse functions $f^{-1}_n,g^{-1}_n$, i.e.

$$\lim_{n\to\infty}\int_K |f_n^{-1}(x)-g_n^{-1}(x)| \mathrm d x =0?$$

If not, is there at least a weaker form of convergence? Thanks for the answer :)

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    $\begingroup$ It does not, take e.g. $f_n(x)=x/n^2+1/n$ and $g_n(x)=x/n^2$ $\endgroup$ – Pietro Majer Jan 24 '18 at 9:22
  • $\begingroup$ Ops, i missed a condition... what if the limit of $f_n$ and $g_n$ is surjective? Thanks for the answer. $\endgroup$ – Martin Jan 24 '18 at 9:25
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The statement is in fact true if we assume that $f_n$ and $g_n$ are bijective and continuous functions form $\mathbb{R}$ to $\mathbb{R}$, hence monotone, say increasing, and converge point-wise to a surjective function $f$ (hence monotone and continuous).

Indeed let $K\subset [a,b]\subset\mathbb{R}$. Since $f$ is surjective, there are $A\in\mathbb{R}$ and $B\in\mathbb{R}$ such that $f(A)<a$ and $f(B)>b$ and by the point-wise convergence, there holds $f_n(A)<a$, $g_n(A)<a$, $f_n(B)>b$, $g_n(B)>b$ eventually as $n\to\infty$. Now $\int_a^b|f_n^{-1}(s)-g_n^{-1}(s)|ds$ is the two-dimensional measure of the set $$ \big\{(s,t)\in[a,b]\times\mathbb{R} : \ f_n^{-1}(s)\wedge g_n^{-1}(s)\le t\le f_n^{-1}(s)\vee g_n^{-1}(s)\big\}$$ which is eventually included in the set $$ \big\{(s,t)\in\mathbb{R} \times[A,B] : \ f_n(t)\wedge g_n(t)\le s\le f_n(t)\vee g_n(t)\big\},$$ whose Lebesgue measure is $\int_A^B|f_n(t)-g_n(t)|dt$. Since by assumption the latter converges to $0$, so does $\int_K|f_n^{-1}(s)-g_n^{-1}(s)|ds$.

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Also note that the function $f(x)=\lim f_n(x)$ is increasing, and being surjective, it is also continuous. Also, in these hypotheses $f_n$ and $g_n$ converge uniformly on compact sets, in fact, it would sufficient to assume $f_n\to f$ and $g_n\to f$ in $L^1_{\it loc}$.

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