Fundamental period of a function ($T_0=\min \{T>0:f(x+T)=f(x)\forall x\}$)

The fundamental period is the least positive period of a function, when one exists. For $f(x)=\sin (2\pi x)$ it is one; for $f(x)={\sin }^2(2\pi x)$ it is $1/2$. A periodic function need not have a fundamental period: a constant function has every positive period, and periodic function without a least positive period is a nonconstant exception. Every nonconstant real-valued function that is continuous and periodic on the real line does have a least positive period.

For a real-valued continuous function, the period group of a function is a closed subgroup: if periods $T_n$ tend to $T$, continuity passes $f(x+T_n)=f(x)$ to the limit. If positive periods could tend to zero, integer multiples of those periods would approach any prescribed displacement $y-x$, forcing $f(y)=f(x)$. A nonconstant function therefore has a positive lower bound for its positive periods. Their infimum $T_0$ is itself a period by closedness. Subtracting an integer multiple of $T_0$ from any other period leaves a period in $[0,T_0)$, which must be zero. This proves the stated structure without assuming differentiability.

Define the indicator function of the rational numbers on the real line by $f(x)=1$ for rational $x$ and $f(x)=0$ otherwise. Every rational shift preserves rationality and irrationality, so every nonzero rational number is a period of a function. Conversely, an irrational shift sends $x=0$ from value one to value zero, so it is not a period. The period group of a function is exactly $\mathbb{Q}$, and the positive periods $1/n$ approach zero without a smallest one. The function is nonconstant and discontinuous everywhere; this is why continuity matters in continuous periodic functions have a least positive period.