A local $U(1)$ transformation is given by \begin{equation} f(x) = e^{i\epsilon(x)} \qquad \text{with} \qquad \epsilon(x) \in C^\infty \, . \end{equation}
Why do we require that the functions in the exponent are infinitely differentiable functions? What would go wrong without this restriction?
In the Standard Model, Gauge Symmetries are made local to introduce interactions. For example in the free Dirac Lagrangian: $$\mathcal{L}_D=\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ it is easy to verify that under a $U(1)$ transformation, it is invariant. By making $U(1)$ local, an interaction is brought into play (namely, the interaction with the electromagnetic photon).
In a more complicated theory, namely in the Glashow-Weinberg-Salam model of the electro-weak interaction, the group that is made local is the $SU(2)_L\times U(1)_Y$.
This procedure leads to a new Lagrangian in which there appear lots of new terms. After lots of rearrangements and new definitions of the fields, there appear 1 Complex massive Field (The $W^\pm$ i.e. the charged bosons mediating weak interactions), 1 real massive field (The $Z_0$ i.e. the neutral bosons mediating weak interactions) and 1 real massless field (The $A^\gamma$ i.e. the photon).
These fields are made up of parts which where the old fields introduced in the process of making local the gauge symmetry, and other parts which are the derivatives of the local phase of the local gauge transformations.
In other words for a gauge transformation of $SU(2)_L\times U(1)_Y$: $$\phi(x)\to \phi'(x)=e^{\frac{i}{2}\vec{\sigma}\cdot\vec{\alpha}(x)}e^{\frac{i}{2}\beta(x)}$$ At the infinitesimal level, we identify: $$\vec{\alpha}(x)=-\frac{\vec{\theta}(x)}{v}\quad\quad\quad\beta(x)=\frac{\theta_3(x)}{v}$$ where $v$ characterized the vacuum choosen, and these $\vec{\theta}$(x) are Goldstone Bosons. The Fields $W^\pm,Z_0,A^\gamma$ are defined in function of these Goldstone Bosons (to be precise are defined in function of the derivatives of these $\theta_i(x)$). And since Goldstone Bosons are defined in function of the local parameters of the local gauge transformation, if the local gauge transformation's parameters happen to be hill defined, then our interacting bosons will be hill defined and all the particle physics will be meaningless.
