What is an example of a real-valued function where an absolute maximum is also an absolute minimum?
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1$\begingroup$ A constant one? $\endgroup$– hmakholm left over MonicaMar 6, 2014 at 3:04
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$\begingroup$ What do you mean by "absolute maximum"? Do you mean a global maximum, i.e. an $x$ such that for every $y$ you have $f(x) \geq f(y)$? $\endgroup$– fgpMar 6, 2014 at 3:10
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$\begingroup$ The terms "absolute maximum" and "global maximum" are synonymous as far as I know. $\endgroup$– DavidMar 6, 2014 at 3:15
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$\begingroup$ Thanks, so it is just a constant. I was overthinking this question and wasn't sure if the answer was just trivial. $\endgroup$– BrianMar 6, 2014 at 3:18
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4 Answers
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1$\begingroup$ (+1): "Brevity is the soul of wit" - Hamlet, act 2, scene 2. $\endgroup$– DavidMar 6, 2014 at 3:14
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$\begingroup$ You should see what I did to circumvent the 50 character minimum.. $\endgroup$– MT_Mar 6, 2014 at 3:20
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$\begingroup$ V clever. I guess you could also have done something like this: \overline{\ \ \ \ \ \ \ \ \ \ \ \ } $\endgroup$– DavidMar 6, 2014 at 3:57
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The only functions which have this property are the constant functions.
Why?
Suppose $f(x) \geq f(m)$ for all $x$. Then $x=m$ is the global minimum of $f$. Likewise, if $f(x) \leq f(M)$ for all $x$. Then $x=M$ is the global maximum of $f$. So if $f(M)=f(m)$, we have $f(m) \leq f(x) \leq f(M)=f(m)$ for all $x$. This forces $f(x)=f(m)=f(M)$ for all $x$ so that $f$ is constant.
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Hint. Suppose that the absolute maximum value is $M$. By definition this means that $f(x)\le M$ for every value of $x$. If the same $M$ is the absolute minimum value, this means that. . .
Can you finish this and determine what $f(x)$ is?
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If $a$ is such that $f(a) \leq f(x)$ and $f(x) \leq f(a)$ for all $x$ in the domain of $f$ we have $f(a) \leq f(x) \leq f(a)$ which implies $f(a) = f(x)$, that is $f$ is constant.
