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Consider : $f(x) = 2^{\lfloor x \rfloor - x}$ . We want to find minimum of $f(x)$. I know that $f(x)$ is not continuous and periodic function but what we can say about minimum and maximum of it ? (Its minimum or maximum is global or local ?)

Please Help!

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Clearly, a periodic function cannot have a strict global maximum, since every value on $[0,1)$ repeats on $[1,2)$.

However, you can easily find maximums of $f$ on $[0,1)$ and each maximum on this interval will be one of an infinite family of maximums. Same for minumums.

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    In this case , Is $f$ periodic function ? – 2017-01-06
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    @S.H.W Yes, with a period of $1$. – 2017-01-06
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    So , Can we say that periodic functions in general doesn't have global maximum or minimum ? – 2017-01-06
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    @S.H.W No we can't. We can say periodic functions don't have global **strict** maximums. – 2017-01-06
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    Okay , And also how you can find minimum of this function ? – 2017-01-06
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    @S.H.W Find the minimum on $[0,1]$. It should be easy. – 2017-01-06
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    But it isn't continuous function – 2017-01-06
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    @S.H.W It is on $[0,1)$... – 2017-01-06
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    Can you calculate this value for me and explain step by step? I'm really confused about it. – 2017-01-06
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    @S.H.W On $[0,1)$, what is $[x]$ equal to? So what is $[x]-x$ equal to? So what is $2^{[x]-x}$ equal to? – 2017-01-06
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    Okay , it is $\frac{1}{2}$ but we can't say it as minimum because this point isn't in the function – 2017-01-06
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    @S.H.W What is $\frac12$? What do you mean by the sentence "It is $\frac12$"? What is the "*it*" in the sentence? Please, word yourself so I can understand. – 2017-01-06
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    $x \in [0,1) \to \lfloor x \rfloor = 0 \to f(x) = 2^{-x} \to f(1) = \frac{1}{2}$ . But we know that $1$ isn't in interval and so we can't say "$\frac{1}{2}$ is minimum of function in $[0,1)$ interval" – 2017-01-06
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    @S.H.W Correct. Which means you just proved that $f$ has no minimum, but it has a maximum (which is $1$). Well done! – 2017-01-06
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    Okay , Therefore it doesn't have minimum and just has maximum . Am I right ? – 2017-01-06