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$f:\mathbb R \longrightarrow \mathbb R$ is continuous, no constant and periodic $\implies$f is bounded.

My answer:

Suppose $f$ is unbounded, we have many possibilities, but since $f$ is continuous so it is unbounded

$$ \text{Does not exist} \displaystyle\lim_{x\to\infty} f(x) \text{ or} \displaystyle\lim_{x\to\infty} f(x)=\infty $$

If the limits do not exist how do I?

Suppose that $\displaystyle\lim_{x\to+\infty} f(x)=+\infty$ the remaining cases are treated in an analogous way.

Now $\displaystyle\lim_{x\to+\infty} f(x)=+\infty \iff \forall \epsilon>0 \, \exists \delta>0: x>\delta \implies f(x)>\epsilon$. As $f$ is periodic, let us suppose of period $P$, for $a\in\mathbb R$, $f(a)=f(a+P)=...=f(a+nP), n\in\mathbb N$.

For $\epsilon=2f(a)>0$, $\exists \delta>0, \forall x>\delta \implies f(x)>2f(a)$ but this is a contradiction (isn't it?), because in $]\delta,+\infty[$ has of existing natural $m$ such that $a+mP>\delta$ because $\mathbb N$ is not increased (is it correct?), and in this case, $f(a+mP)=f(a)<2f(a)$. What do you think ?

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  • $\begingroup$ It is bad form to use both "If" and "$\implies$". The symbol takes precedent over the word, so what you have is a premise that has a premise but no conclusion; that is, "(if $f\colon\mathbb{R}\to\mathbb{R}$ is continuous and periodic)" is your incomplete premise. Use either "If...then", or just $\implies$. $\endgroup$ Aug 11, 2022 at 15:03
  • $\begingroup$ You're absolutely right, it was "just" a little rook. Thanks $\endgroup$
    – Pierre
    Aug 11, 2022 at 15:08
  • $\begingroup$ @ArturoMagidin In my book it's not so much bad form as simply incoherent: "If A implies B then..." is not what was meant $\endgroup$ Aug 11, 2022 at 15:19
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    $\begingroup$ Does this answer your question? How do I show that all continuous periodic functions are bounded and uniform continuous? $\endgroup$ Aug 11, 2022 at 16:12
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    $\begingroup$ I just wanted to know if my resolution is correct or not, if possible $\endgroup$
    – Pierre
    Aug 11, 2022 at 17:17

1 Answer 1

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You want to show that $f(\mathbb{R})$ is a bounded set. Now observe that by periodicity, there is a compact interval $[a,b]$ such that $f(\mathbb{R}) = f([a,b])$. What do we know about the image of compact sets under continuous functions?

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  • $\begingroup$ Are limited. What I wanted to do in my answer was not to use compactness, just initial calculus, so to speak $\endgroup$
    – Pierre
    Aug 11, 2022 at 15:07
  • $\begingroup$ Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here $\endgroup$ Aug 11, 2022 at 16:14
  • $\begingroup$ I just wanted to know if my resolution is correct or not, if possible $\endgroup$
    – Pierre
    Aug 11, 2022 at 17:17

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