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Let $X$ be a metric space, $f_n: X \to X$, $g_n: X \to X$,

$f_n(x) \to f(x)$, $g_n(x) \to g(x)$ ($n \to \infty$).

Is $f_n(g_n(x)) \to f(g(x))$ ?

Here: 1) pointwise convergence;

2) uniform convergence.

So, there are 2 cases in my question.

In the first case $f_n \to f$ and $g_n \to g$ pointwise.

In the second case $f_n \to f$ and $g_n \to g$ uniformly.

Thank you very much!

Dina
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Dina
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1 Answers1

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No to both. Here are two examples:

1) Take $f_n(x)=x^n$ and $g_n(x)=1-\frac{x}{n}$ on the real interval $[0,1]$, then they converge pointwise to $f=\begin{cases}1&x=1\cr 0&x<1\end{cases}$ and $g=1$ respectively, so $f(g(x))=1$, but $f_n\circ g_n(x)\to e^{-x}$.

1+2) $f_n(x)=f(x)=\begin{cases}1&x>0\cr 0&x\le0\end{cases}$ and $g_n(x)=x/n$. They converge uniformly to $f$ and $g=0$ on $[-1,1]$ and $f\circ g(x)=0$; yet $f_n\circ g_n=f$.

Chrystomath
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    Would there be a condition to put into $f_n$ and $g_n$ to make it converge pointwise/uniformly? Inversability of $f_n$? – Marco Aguiar Jun 05 '16 at 06:02
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    @MarcoAguiar Continuity of the limit function $f$ is needed. This, together with uniform convergence of $f_n$ and pointwise convergence of $g_n$ lead to $f_n(g_n) \to f(g)$. – Alecos Papadopoulos Apr 30 '17 at 18:17
  • Further, if we include the continuity of $f$ and $g$, then what we can say about the same? – MANI Aug 30 '19 at 12:09