prove a function to be constant

Let the function $f\colon \mathbb{R} \to\mathbb{R}$ have arbitrarily small positive periods, in the sense that if $\delta>0$ then there exists $T \in (0,\delta)$ such that $f(t + T) = f(t)$ for all $t \in \mathbb{R}$.

(i) Prove that if $f$ is continuous then $f$ is constant.

(ii) What if $f$ is not assumed to be continuous?

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Pippo, what methods have you tried for this problem? For part (ii), consider using arbitrarily small periods "T" that are rational numbers. Does it limit what "f" can be on the irrationals? – Conan Wong Nov 27 '12 at 23:07

(ii) is answered by $f(x)=\mathbf 1_\mathbb Q(x)=\begin{cases}1&x\in\mathbb Q\\0&x\notin\mathbb Q\end{cases}$.

(i) Let $x\in\mathbb R$ be arbitrary. Let $\epsilon>0$ be given. There exists a $\delta>0$ such that $|f(y)-f(x)|<\epsilon$ if $|y-x|<\delta$. Let $T\in(0,\delta)$ be a period. Then $f(nT)=f(0)$ for all $n\in \mathbb Z$ (by induction on $n$). Let $n=\left\lfloor\frac xT\right\rfloor$. Then $nT\le b<(n+1)T$ and hence $|b-nT|<T<\delta$. We conclude $|f(x)-f(0)|=|f(x)-f(nT)|<\epsilon$. Since $\epsilon$ was arbitrary, $f(x)=f(0)$. Since $x$ were arbitrary, $f$ is constant.

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(i) For each $n$, there is $0<T_n<n^{—1}$ such that $f(x+T_n)=f(x)$ for $x\in \Bbb R$. We can show that $f$ is uniformly continuous on $[0,T_n]$. So fix $\varepsilon>0$, and $n$ such that if $x\in [0,T_n]$, then $|f(x)-f(0)|<\varepsilon$. Let $x\in \Bbb R$. Write it as $N_nT_n+c_n$, where $N_n$ is an integer and $c_n\in (0,T_n)$. Then $$|f(x)-f(0)|=|f(c_n)-f(0)|\leqslant \varepsilon.$$ As it's true for all $x$ and all $\varepsilon$ we get that $f(x)=f(0)$ for all $x\in\Bbb R$.

(ii) A counter-example is the characteristic function of rational numbers.

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Let $\Delta = \{\tau | f(0) = f(\tau)\}$. By assumption, there exists a sequence $T_i \downarrow 0$ such that $f$ is $T_i$-periodic. Hence $\{kT_i\}_{k \in \mathbb{Z}} \subset \Delta$ for all $i$. It follows from this that $\Delta$ is dense.

If $f$ is not continuous, then $1_\mathbb{Q}$ shows $f$ need not be constant.

If $f$ is continuous, then $\Delta$ is closed (and dense), hence $\Delta = \mathbb{R}$.

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