I am solving CMI PHD questions.One of the questions from real analysis is as follows:
Question
Let $f:\mathbb {R\to R}$ be a function such that $f(x+1)=f(x)$ for all $x\in \mathbb R$,then which of the following are true:
$(a) f$ is bounded.
$(b)f$ is bounded if it is continuous.
$(c)f$ is differentiable if it is continuous.
$(d)f$ is uniformly continuous if it is continuous.
My answer to this question is as follows:
Take $f(x)=\sec(2\pi x)$ and assign $0$ value to points where it is undefined to show that $(a)$ is false.Then $(b)$ is true because $f$ is $1$-periodic and hence $f(\mathbb R)=f([0,1])$ which is bounded if $f$ is continuous.For disproving $(c)$ take the function $g(x)=\frac{1}{2} |x|$ on $[\frac{1}{2},\frac{1}{2}]$ and replicate it periodically on $\mathbb R$.Finally $(d)$ is true because any periodic continuous function is uniformly continuous.
Is my solution correct?