Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a periodic function (not necessarily continuous). The fundamental period of $f$ is defined by: $$T^{*}=\inf\{T>0:T\hspace{0.1cm} \mbox{is a period of}\hspace{0.1cm} f , \hspace{0.1cm} \mbox{i.e.},\hspace{0.1cm} f(x+T)=f(x), \,\,\,\, \forall\,x \in \mathbb{R}\}.$$ Then show that:
i) If $\hspace{0.1cm}T^{*}\neq 0$ then $T^{*}$ is a period of $f$.
ii) If $\hspace{0.1cm}T^{*}=0$ and $f$ is continuous then $f$ is constant.
This is an attempt of i):
Let $A=\{T>0:T\hspace{0.1cm} \mbox{is a period of}\hspace{0.1cm} f , \hspace{0.1cm} \mbox{i.e.},\hspace{0.1cm} f(x+T)=f(x), \,\,\,\, \forall\,x \in \mathbb{R}\}$. Since $T^{*}=\inf A$ then or $T^{*}\in A$ or $T^{*}\notin A$.
Suppose that $T^{*}\notin A$.
If existe $\varepsilon >0$ tal que $\,\big(B_{\varepsilon}(T^{*})-\{T^{*}\}\big)\cap A = \emptyset$$\hspace{0.1cm}$ then$\hspace{0.1cm}$ $c=T^{*}+\varepsilon/2$ meets that $T^{*}<c<T_n$, $\forall n\in \mathbb{N}$. This contradicts that $T^{*}=\inf A$.
Otherwise $\,\,\,\forall\, \varepsilon >0$, $\,\big(B_{\varepsilon}(T^{*})-\{T^{*}\}\big)\cap A\neq \emptyset$. Then $T^{*}$ is a point of accumulation of $A$ then there is a sequence $\{T_n\}_{n\in \mathbb{N}}\subset A$ such that $T_n$ converges to $T^{*}$. $\,\,T_n \in A$ then $f(x+Tn)=f(x)$, $\forall x \in \mathbb{R}$. ( If $f$ were continuous, we would have that $f(x+T^{*})=f(x)$ and therefore $T^{*}\in A$¡! ).