Vtyjkyuk

Here is Prob. 14, Sec. 5.4, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

A function $f colon mathbbR to mathbbR$ is said to be periodic on $mathbbR$ if there exists a real number $p > 0$ such that $f(x+p) = f(x)$ for all $x in mathbbR$. Prove that a continuous periodic function on $mathbbR$ is bounded and uniformly continuous on $mathbbR$.

Here is another Mathematics Stack Exchange post on this very problem.

In this post, the boundedness part is clear. Here is my presentation thereof.

As $f$ is continuous on the closed bounded interval $[0, p]$, so $f$ is bounded on this interval, by virtue of Theorem 5.3.2 in Bartle & Sherbert. So there exists a real number $M > 0$ such that $$ lvert f(x) rvert < M qquad mbox for all x in [0, p]. $$

Now if $x$ is any real number, then since $p > 0$, we can find a natural number $n$ such that $np > x$; let $N$ be the smallest such natural number. Then $$ Np > x geq (N-1)p. $$
So $$ p > x - (N-1)p geq 0, $$ and therefore
$$ biglvert f big( x-(N-1)p big) bigrvert < M. $$
As $f$ is periodic with period $p$, so we must have
$$ f(x) = f big( x-(N-1)p big), $$
which implies that
$$ lvert f(x) rvert = biglvert f big( x-(N-1)p big) bigrvert < M. $$

Hence $$ lvert f(x) rvert < M qquad mbox for all x in mathbbR. $$
So $f$ is bounded on $mathbbR$.

Is this proof correct and any clearer?

Now for the uniform continuity of $f$!!

Let us take any real number $varepsilon > 0$.

As $f$ is continuous on the closed bounded interval $[0, 2p]$, so $f$ is uniformly continuous on this interval, by Theorem 5.4.3 in Bartle & Sherbert. So there exists a real number $delta > 0$ (and this $delta$ depends only on our $varepsilon$) such that
$$ lvert f(x) - f(u) rvert < varepsilon $$
for any points $x, u in [0, 2p]$ such that
$$ lvert x-u rvert < delta. $$
Let us choose our $delta$ such that $delta < p$.

Now let $x, y in mathbbR$ such that $lvert x - y rvert< delta$.

As $2p > 0$, so we can find natural numbers $m$ and $n$ such that $2pm > x$ and $2pn > y$; let $M$ and $N$ be the smallest such natural numbers. Then we must have
$$ 2pM > x geq 2p(M-1) qquad mbox and qquad 2pN > y geq 2p(N-1), $$
and so
$$ 2p > x - 2p(M-1) geq 0 qquad mbox and qquad 2p > y - 2p(N-1) geq 0. $$
Since $f$ is periodic with period $p$, we also have
$$ f(x) = fbig( x - 2p(M-1) big) qquad mbox and qquad f(y) = fbig( y - 2p(N-1) big). $$

Now if we could show that the $M$ and the $N$ postulated above must be equal, then
we must have
$$ leftlvert big( x-2p(M-1) big) - big( y-2p(N-1) big) rightrvert = lvert x-y rvert < delta, $$
and also both $x-2p(M-1)$ and $y-2p(N-1)$ are in the interval $[0, 2p]$ (in fact the interval $[0, 2p)$). Therefore we obtain
$$ lvert f(x) - f(y) rvert = leftlvert fbig( x-2p(M-1) big) - f big( y-2p(N-1) big) rightrvert < varepsilon, $$
from which it follows that $f$ is uniformly continuous on $mathbbR$.

But how to show that the $M$ and the $N$ must be equal? Or, is this the way pointed out in one of the answers to the question here?