a set of functions that are pointwise equicontinuous but not uniformly equicontinuous, supposing the domain of f is noncompact

Question

Can anyone think of an example of such set of functions?(If domain is compact then pointwsie equicontinuity implies uniformly equicontinuous)

Answer 1

No problem. Let $\Omega = (0,1)$, and for $n \in \mathbb{Z}^+$,

$$f_n(x) = \begin{cases}1 - n\cdot x &, x < \frac{1}{n}\\\quad 0 &, x \geqslant \frac{1}{n}. \end{cases}$$

Every point has a neighbourhood on which all but finitely many of the $f_n$ vanish identically, hence the family is equicontinuous: Let $\varepsilon > 0$ be given. For any $x \in (0,1)$, let $N(x) = \lfloor \frac{2}{x}\rfloor$, and $\delta_{1,x} = \frac{x}{2}$. For $n > N(x)$, $f_n$ vanishes identically on $(\frac{x}{2},1)$. For $n \leqslant N(x)$, $f_n$ is Lipschitz continuous with Lipschitz constant $n$, hence for

$$\delta_x = \min \left\{\delta_{1,x}, \frac{\varepsilon}{N(x)}\right\},$$

we have $\lvert y-x\rvert < \delta \implies \lvert f_n(y) - f_n(x)\rvert < \varepsilon$.

But $f_n(\frac{1}{2n}) - f_n(\frac{1}{n}) = \frac{1}{2}$ and $\lvert \frac{1}{n} - \frac{1}{2n}\rvert = \frac{1}{2n}$ for all $n$, so the family is not uniformly equicontinuous.