$f$ is continuous, no constant and periodic $\implies$ f is bounded

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

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 ?

• 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 $\implies$. Aug 11, 2022 at 15:03
• You're absolutely right, it was "just" a little rook. Thanks Aug 11, 2022 at 15:08
• @ArturoMagidin In my book it's not so much bad form as simply incoherent: "If A implies B then..." is not what was meant Aug 11, 2022 at 15:19
• Does this answer your question? How do I show that all continuous periodic functions are bounded and uniform continuous? Aug 11, 2022 at 16:12
• I just wanted to know if my resolution is correct or not, if possible Aug 11, 2022 at 17:17

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?