If the solution is correct could someone please explain to me why it's correct with some detail

(J (Ink) ) k=1] Q14. A function f : R - R is said to be periodic on R if there exists a number p > 0 s.t. f(x + p) = f(x) for all r E R. Prove that a continuous periodic function on R is bounded and uniformly continuous on R. Solution. It is easy to deduce f(x) = f(x + kp) for all r E R and for all ke Z. Note that sup f(x)| = sup If(x)| TER TE[0.p] Because [0, p] is a closed and bounded interval, so f is bounded on [0, p] and hence on R. We need to show f is uniformly continuous on R. Let ( > 0. Because f is uniformly continuous on [-p, 2p], there is o > 0s.t. for allx, y e [-p. 2p]. (x-y)