Prove that a periodic continuous function on $\Bbb{R}$ is bounded and uniformly continuous on $\Bbb{R}$.
Proof:
First I show that $f$ is bounded on $[0,p]$. By way of contradiction, suppose that it is unbounded on the interval. Then for every $n \in \Bbb{N}$, there exists an $x_n \in [0,p]$ such that $|f(x_n)| > n$. Thus $\{|f(x_n)|\}$ is a increasing sequence of positive numbers diverging to infinity, and therefore every subsequence must diverge to infinity. However, $x_n \in [0,p]$ for every $n$ implies that there exists a convergent subsequence $x_{k_n}$ converging to $L$, which must be in $[0,p]$ as it is closed. By continuity of $f$, $f(x_{k_n}) \to f(L)$, and by continuity of the absolute value function, $|f(x_{k_n})| \to |f(L)| < \infty$, which is a contradiction.
Now we show that $f([0,p]) = f([n,(n+1)p])$ for every $n \in \Bbb{Z}$. Note that $y \in f([n,(n+1)p])$ iff there exists an $x$ such that $n \le x \le (n+1)p$ or $0 \le x - np \le p$ if and only if $y = f(x) = f(x-np) \in f([0,p])$.
Now we show that $f$ is bounded on $\Bbb{R}$. Since $[0,p]$ is compact, and $f$ is continuous, $f$ attains an minimum and maximum; i.e., $f([0,p])$ has a largest and smallest element. By the above result, the minimum and maximum is the same for $f([np,(n+1)p])$ for every $n \in \Bbb{Z}$. Let $K$ denote the maximum of the absolute value of these extrema. Hence, $|f(x)| \le K$ for every $x \in [0,p]$. Now, let $x \in \Bbb{R}$. There exists an integer $n$ such that $ n \le \frac{x}{p} \le n+1$ or $0 < x-np < p$. Therefore $|f(x)| = |f(x-np)| \le K$.
Here is the part that's giving me trouble, namely, showing that $f$ is uniformly continuous on $\Bbb{R}$. Since $[0,p]$ is compact, and $f$ is continuous, $f$ must also be uniformly continuous on this set. Let $x,y \in \Bbb{R}$. Then there exist integers $n$ and $k$ such that $0 \le x - np \le p$ and $0 \le y - kp \le p$. I I am trying to find a $\delta' > 0$ such that $|x-y| < \delta'$ implies $|(x-np)-(y-kp)| < \delta$, where $\delta > 0$ is the number we get from $f$ being uniformly continuous on $[0,p]$. Then I would use periodicity to conclude that $|f(x) - f(y)| = |f(x-np)-f(y-kp)| < \epsilon$, but obviously $\delta $ cannot depend upon $n$ and $k$, since it would then depend upon $x$ and $y$.
I could use a hint. At any rate, does the first part of my proof seem correct?