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I need to check whether $g(x)$ is also continuous for: $$ g(x)=\frac{1}{\min\{f(x),-1\}} $$ Two questions:

  1. Can I show the continuity of $g(x)$ by using $f(x)=x$ or should I use more generalized approach?
  2. Am I right that $g(x)$ will be continuous, as composition of two continuous functions ($min\{\}$ and $f(x)$) is also continuous and $\frac{1}{x}$ is continuous for $(-\infty,-1)$ ?
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Can't you just re-write it as a piecewise function, and show continuity from that? –  Maz Feb 16 '13 at 16:46
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2 Answers

up vote 4 down vote accepted

You'll need to be more general. If you know that $\min\{f(x),h(x)\}$ is continuous whenever $f(x),h(x)$ are, then $\min\{f(x),-1\}$ is a continuous non-zero function, and so $g$ is the reciprocal of a non-zero continuous function, so continuous.

As a hint for how to show that $\min\{f(x),h(x)\}$ is continuous if $f(x),h(x)$ are, note that we have $$\min\{a,b\}:=\frac{a+b-|a-b|}2$$ for any real $a,b.$ (Why?) Thus, $$\min\{f(x),h(x)\}=\frac{f(x)+h(x)-|f(x)-h(x)|}2.$$ Some basic continuity results should let you prove the desired result.

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  1. No. When proving continuity, the function $f$ is an arbitrary, continuous function. You could have chosen $f$ to be whatever you like, were you trying to generate a counter example.
  2. Although the intuition is right, your statement is weird. For example, how exactly would you define the function $\min$? And how would you prove it's continuous? Also, it doesn't help you much saying that $\frac{1}{x}$ is continuous on $(-\infty, -1)$, since you have to prove continuity $\forall x\in \mathbb{R}$.

Finally, as @Cameron Buie suggested, you might want to try a more general approach. Start with proving (by definition, even) that $\min{\{f(x),-1\}}$ is continuous. Then show that $\min{\{f(x),-1\}}\neq 0$ $\forall x\in \mathbb{R}$. And then conclude the continuity of $g(x)$.

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Ad 2. I mentioned continuity of $\frac{1}{x}$ for $(-\infty,-1)$, as $g(x)$ is essentially $\frac{1}{x}$ for this range (and a constant function $-1$ for the $(-1,\infty)$). Based on your answer, it looks like I don't even need to use that. –  REACHUS Feb 16 '13 at 19:07
Right. Because you can always use the fact that if $h(x),s(x)$ are continuous on $I$ and $s(x)\neq 0$ for all $x\in I$, then the function $\frac{h(x)}{s(x)}$ is continuous on $I$. –  Ludolila Feb 16 '13 at 19:11
If you're curious about how we can explicitly define $\min$ (at least, on sets $\{a,b\}$ with $a,b\in\Bbb R$), see my updated post. That will actually let us explicitly define $g(x)$ without any set notation. –  Cameron Buie Feb 18 '13 at 0:04
@CameronBuie, that's great :) Now I see the elegant way of proving the continuity of $\min\{ \}$ (as opposed to the $\varepsilon, \delta$ thing I was doing bef0re...) –  Ludolila Feb 18 '13 at 0:14
Glad I could help you, too! –  Cameron Buie Feb 18 '13 at 0:17
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