Need to prove $f(x+P)=f(x)$ for $P>0$ Let $f$ be continuous on $\mathbb R$ and suppose that there exists a number $P > 0$ such that $f(x + P) = f(x)\ \forall\ x \in \mathbb R.$  Prove that the function is bounded and uniformly continuous on $ \mathbb R.$  
Thus far it is readily apparent that once I am able to determine the boundedness of the function, it follows that the function is uniformly continuous.  My thought is to choose $f(x) = \cos(x),$ and choose $P = 2n\pi,n\in \mathbb Z.$ 
Looking for assistance on how to best present the proof if this is a viable avenue to answer the question.
 A: Hint: use the fact that if a function is continuous on a compact set, then it is uniformly continuous there. Consider, for example, the interval $I=[-k,\, k]$. It is immediate the $I$ is compact, and therefore $f$ is uniformly continuous there. If $x \not \in I$, then use the fact that $f$ is periodic.
A: Break down the Real line into the compact intervals $[KP, (K+1)P]$ ; $ K \in \mathbb Z$. Then f is uniformly continuous in each of these. Take the largest $\delta$ for a given $\epsilon$ in any interval as above . This $\delta$ will work for any and all intervals, except possibly at the endpoints. 
Let $x,y$ be in $[KP, (K+1)P]$ for some integer $K$. Then, for fixed $\epsilon$ by uniform continuity (from compactness and continuity), $|f(x)-f(y)|< \epsilon $ when $|x-y|< \lambda$ for fixed $\lambda$. If $x,y$ are not in the same period, use the triangle inequality to take care of any issue. It follows that for fixed $\epsilon$, there exists a fixed $\delta (\epsilon)$ so that $|f(x)-f(y)|< \epsilon $ when $|x-y|< \delta (\epsilon)$.
