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If $f,g$ are continuous at $a$, show that $h(x)=\max(f(x),g(x))$ and $k(x)=\min(f(x),g(x)) \text{ are also continuous } \forall x\in X$.

Hello. I am aware this question has been asked before and I also know the solution. But since I did it another way, I wanted to make sure my reasoning is correct. I'll make my argument just for one part since it is analogous:

We know $f,g$ are continuous at $a$. Suppose $h(x)$ is not continuous at $a$. This would imply that either $f$ or $g$ are not continuous there, which is a contradiction.

Is this ok?

Robert Shore
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H44S
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  • It's not okay because it's an insufficient explanation of your reasoning. Why would $h(x)$ not continuous at $a$ imply that either $f$ or $g$ is not continuous there? – Robert Shore Oct 16 '19 at 23:42
  • Because $f(a) \geq g(a)$ or $f(a) \leq g(a)$, so $h$ would be whichever one is greater, isn't it? Either way, it would have to be continuous there, as it is defined that way. – H44S Oct 16 '19 at 23:48
  • The question wants you to prove that it's defined that way – kingW3 Oct 17 '19 at 00:09
  • Why can't $h$ "switch" from $f$ to $g$ right at $a$? If I were grading this explanation, I wouldn't accept it. Something like the answer below is much more rigorous. You need to use either the definition of continuity or some properties of continuity that have already been proved. The answer below takes the latter approach: The composition of continuous functions is also continuous, as are their sum, their difference, their quotient (assuming the denominator is non-zero), and their absolute value. – Robert Shore Oct 17 '19 at 00:10
  • I see. I was just thinking that it wouldn't matter if h switched from one function to another, as you said, since they are both continuous at a. – H44S Oct 17 '19 at 00:16

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Note that $$\max \{f,g \} = |(f+g)/2|+|(f-g)/2|$$ and $$\min \{f,g \} = |(f+g)/2|-|(f-g)/2|$$

Since all function involved are continuous the resulting functions are also continuous.

Mohammad Riazi-Kermani
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