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Show that if f is continuous and periodic then f attains both its minimum and its maximum.

The solution is given below:

enter image description here

But I wonder why he choose the k like this and is the solution does not contain any flaws?

I can not understand the general idea of the proof, could anyone explain it for me please?

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  • $\begingroup$ The proof seems unnecessarily involved to me. I would just observe that by periodicity, the image is $f(I)$ for some closed interval $I=[a,b]$, recall that the continuous image of a compact connected set is compact and connected, and that the only compact connected sets in $\mathbb R$ are closed (bounded) intervals. Thus, the image is of the form $[f(c),f(d)]$ for some $c,d\in I$. $\endgroup$ – MPW Oct 29 '18 at 8:55
  • $\begingroup$ I can not use this concepts until now ........ I need to see if the above proof is consistent or no @MPW $\endgroup$ – hopefully Oct 29 '18 at 8:59
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Consider the picture below:interval

We first divide the real line into the intervals $[0,T]$, $[T,2T]$,... (negative also but we assume that $C>0$ for this picture). Suppose that $C$ exists such that $f(C)>M = \sup\{f(x):0\leq x \leq T\}.$

We pick $k$ such that $C\in [kT,(k+1)T]$. This is possible since these intervals cover $\mathbf{R}$. This is precisely what the author means by letting $k = \lfloor C/T \rfloor $ that is the largest integer $k$ such that $k\leq C/T$ which implies that $k\leq C/T\leq k+1$ and therefore $C\in [kT,(k+1)T].$

Now $kT\leq C\leq (k+1)T\Rightarrow 0\leq C-kT\leq (k+1)T-kT = T$ and therefore this would imply that if we write $C' = C-kT$

$$f(C') = f(C-kT) = f(C) >M$$

However $C'\in [0,T]$ by the above inequality and therefore $f(C')\leq M$ thus we have derived a contradiction.

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  • $\begingroup$ but why he said contradicting that M is maximum ...... was it assumed that Mis maximum (I know that it was assumed that it is supremum not maximum)? $\endgroup$ – hopefully Oct 29 '18 at 8:45
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    $\begingroup$ $f$ restricted to $[0,T]$ is continuous and $[0,T]$ is compact. What do you know about maximum and supremum of continuous functions on compact sets? $\endgroup$ – Olof Rubin Oct 29 '18 at 8:47
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    $\begingroup$ okay thank you :) $\endgroup$ – hopefully Oct 29 '18 at 8:51
  • $\begingroup$ what will be the value for k if I want to proof hat f attains its minimum? $\endgroup$ – hopefully Oct 29 '18 at 9:31
  • $\begingroup$ just repeat the proof exchanging supremum with infimum and maximum with minimum $\endgroup$ – Olof Rubin Oct 29 '18 at 9:33
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This proof, while correct, is very clumsy, to say the least.

Since $f$ is periodic any value taken by $f$ is already taken in the interval $[0,T]=:J$. The restriction of $f$ to $J$ is continuous, hence $f\restriction J$ takes a maximal value $M$ at some point $\xi\in J$. It follows that $\max_{x\in{\mathbb R}} f(x)=M=f(\xi)$.

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