a set of functions that are pointwise equicontinuous but not uniformly equicontinuous, supposing the domain of f is noncompact Can anyone think of an example of such set of functions?(If domain is compact then pointwsie equicontinuity implies uniformly equicontinuous)
 A: 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.
A: For a simpler example, take a set of functions consisting of a single function.  In that case, the set being pointwise equicontinuous corresponds to the function being continuous.  And the set being uniformly equicontinuous corresponds to the function being uniformly continuous.
So any continuous function that is not uniformly continuous will do the job.  For example on $\mathbb{R}$ take $f(x) = x^2$, which is not uniformly continuous since its derivative is unbounded.
