I am reading from the book Multidimensional real analysis vol I by Duistermaat and Kolk and am trying to understand the following theorem from it:

Theorem 1.4.2: Suppose $A\subset \mathbb R^n,B\subset \mathbb R^p, f:A\to \mathbb R^p, g:B\to \mathbb R^q, a\in\overline{A},b\in\overline{B},c\in\mathbb R^q$. Further suppose $\lim_{x\to a}f (x)=b $ and $\lim_{y\to b}g(y)=c$. Then $\lim_{x\to a}(g\circ f )(x)=c$.

My question is somewhat silly. For $\lim_{x\to a}(g\circ f )(x)$ to make sense we must have $a\in\overline{\mbox {dom }g\circ f}$, i.e. we must have $a\in\overline{A\cap f^{-1}(B)}$. However we are only given $a\in\overline{A}$. Is this an error ?


You're right. This seems to be an error. You can repair it by restricting yourself to $$f:A\rightarrow B.$$ Without it, $g\circ f$ may not be well defined.

Another thing. You stated $$\lim_{x\rightarrow a} (g\circ f)(x)=b.$$ Shouldn't this $b$ be a $c$?

  • $\begingroup$ Thanks; corrected. $\endgroup$ – Shahab Aug 15 '18 at 13:59
  • $\begingroup$ Why do you feel that $g\circ f $ may not be well defined? The function with an empty domain is a function: math.stackexchange.com/a/789411/10575 $\endgroup$ – Shahab Aug 16 '18 at 1:59
  • $\begingroup$ Can I say $a\in\overline{A\cap f^{-1}(B)}$ to repair the error? $\endgroup$ – Shahab Aug 16 '18 at 2:02
  • $\begingroup$ Yes of course $g\circ f:\emptyset\rightarrow\emptyset$ is defined, but if you want to evaluate $\lim_{x\rightarrow a}g\circ f$ you need to know what $g\circ f$ is at least for a sequence converging to $a$. $\endgroup$ – humanStampedist Aug 16 '18 at 7:51
  • $\begingroup$ $a\in \overline{A\cap f^{-1}(B)}$ should also work, because then you have a sequence $x_n\in A\cap f^{-1}(B)$ with $x_n\rightarrow a$. Also $f(x_n)\in B$, hence $g\circ f(x_n)$ is defined. $\endgroup$ – humanStampedist Aug 16 '18 at 8:01

If $f:[0,1[\rightarrow \mathbb R$ is defined by $f(x)=x$, then we can say $\lim_{x\to 1}f (x)=1 $, even if $1$ is not contained neither on the demain or the image of $f$. The need for $a$ to be in be in ${\overline{ f^{-1}(B)}}$ may exist but is implicitly asserted by $\lim_{x\to a}f (x)=b $.

Also, the corret result should be: $\lim_{x\to a}(g\circ f )(x)=c$.

  • $\begingroup$ Shouldn't it be possible that $g\circ f$ is not defined. For Example $g(x)=\sqrt{-x-1}$ and $f(x)=x^2$. Then $g\circ f(x)=\sqrt{-x^2-1}$, which is not well defined in the real numbers. $\endgroup$ – humanStampedist Aug 15 '18 at 13:51
  • 2
    $\begingroup$ I think it might even be more extreme. Let $f:\mathbb{R}\rightarrow(0,\infty)$ and $g:(-\infty,0)\rightarrow \mathbb{R}$ with $\lim_{x\rightarrow 0}f(x)=0$ and $\lim_{x\rightarrow 0}g(x)$ exists. Then you cannot define $\lim_{x\rightarrow 0} g\circ f(x)$, since $g\circ f$ is not well defined for any $x$, although $0\in \overline{(-\infty,0)},\overline{(0,\infty)}$. $\endgroup$ – humanStampedist Aug 15 '18 at 14:03
  • $\begingroup$ Both good points, I assumed the limit of $g\circ f$ would exist to begin with. $\endgroup$ – Mefitico Aug 15 '18 at 15:54

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