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

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 $$ \big\lvert f \big( x-(N-1)p \big) \big\rvert < M. $$ As $f$ is periodic with period $p$, so we must have $$ f(x) = f \big( x-(N-1)p \big), $$ which implies that $$ \lvert f(x) \rvert = \big\lvert f \big( x-(N-1)p \big) \big\rvert < M. $$

Hence $$ \lvert f(x) \rvert < M \qquad \mbox{ for all } x \in \mathbb{R}. $$ So $f$ is bounded on $\mathbb{R}$.

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 $x, y \in \mathbb{R}$ 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 $$ f(x) = f\big( x - 2p(M-1) \big) \qquad \mbox{ and } \qquad f(y) = f\big( y - 2p(N-1) \big). $$

Now if we could show that the $M$ and the $N$ postulated above must be equal, then we must have $$ \left\lvert \big( x-2p(M-1) \big) - \big( y-2p(N-1) \big) \right\rvert = \lvert x-y \rvert < \delta, $$ 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 $$ \lvert f(x) - f(y) \rvert = \left\lvert f\big( x-2p(M-1) \big) - f \big( y-2p(N-1) \big) \right\rvert < \varepsilon, $$ from which it follows that $f$ is uniformly continuous on $\mathbb{R}$.

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?

  • $\begingroup$ Nice problem and solution (+1). $\endgroup$
    – user798113
    Jun 7, 2021 at 5:51

1 Answer 1


The proof of boundedness on ${\mathbb R}$ seems correct but can be made considerably shorter. As a general guideline, I recommend preferring prose to notation as much as possible.

Having proved that $f(x)$ is bounded in modulus by $M$ on the interval $[0, p]$, consider, as a first case, a point $a > p$ in ${\mathbb R}$. Since ${\mathbb R}$ is an Archimedean field, there exists the largest integer $N$ such that $Np < a < (N+1)p$. It follows that $$ Np - a \in [0, p], $$ so $f(a) = f(a - Np)$. Therefore, $|f(a)| \leq M$. The proof for the case $a < 0$ is analogous.

As for uniform continuity, given an $\epsilon > 0$, the same $\delta_{\epsilon} > 0$ will satisfy the condition $$ |x_{1} - x_{2}| < \delta_{\epsilon} \quad \mbox{ implies } \quad |f(x_{1}) - f(x_{2})| < \epsilon $$ on every subinterval $[N p, (N+1) p]$ of ${\mathbb R}$ (where $N$ ranges over the integers). And if the points $x_{1}, x_{2}$ have between them an integral multiple $K p$ of $p$, then $$ |x_{1} - K p| \leq |x_{1} - x_{2}| < \delta_{\epsilon} \mbox{ and } |x_{2} - K p| \leq |x_{1} - x_{2}| < \delta_{\epsilon}, $$ so $$ |x_{1} - x_{2}| \leq |x_{1} - K p| + |x_{2} - K p| < 2\delta_{\epsilon}. \quad \mbox{(the first inequality is the triangle inequality.)} $$ Therefore, the condition $$ |x_{1} - x_{2}| < 2 \delta_{\epsilon} \quad \mbox{ implies } \quad |f(x_{1}) - f(x_{2})| < 2 \epsilon $$ holds on the union of the subintervals $[N p, (N+1) p]$; i.e., on ${\mathbb R}$.


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