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

Martin Sleziak
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user3404321
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2 Answers2

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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.

Daniel Fischer
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    Yes, every point has a neighborhood such that fn(x)=0 except for finitely many fn but definition of pointwise equicontinuity requires all fn to be satisfied. And isn't this set of function uniformly equicontinuous? – user3404321 Mar 12 '14 at 22:51
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    Every finite family of continuous functions is equicontinuous, so since there is a neighbourhood $V$ of $x$ on which all but finitely many vanish identically, the family is equicontinuous (for any $\varepsilon > 0$, there is a neighbourhood $U$ of $x$ with $\lvert f_k(y) - f_k(x)\rvert < \varepsilon$ for the finitely many $f_k$ that don't vanish identically on $V$; choose $U\cap V$). The family is not uniformly equicontinuous, since there are arbitrarily close points where one (or more) of the functions differ by $\frac12$. – Daniel Fischer Mar 12 '14 at 22:57
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    I see it now. Thanks a lot! – user3404321 Mar 12 '14 at 23:10
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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.

PatrickR
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