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Let $X,Y$ be metric spaces and $(f_n)_n$ a family of functions $X \rightarrow Y$.

We say that $(f_n)_n$ is equicontinuous if $\forall x\in X \quad \forall \varepsilon >0 \quad \exists \delta >0 \quad | \quad \forall n \quad \forall y\in X \quad (d(x,y)< \delta \implies d(f_n(x),f_n(y)) < \varepsilon)$.

I would like to know whether this along with uniform continuity of each $f_n$ impies that $(f_n)_n$ is uniformly equicontinuous, that is : $\forall \varepsilon >0 \quad \exists \delta >0 \quad | \quad \forall n \quad \forall x,y\in X \quad (d(x,y)< \delta \implies d(f_n(x),f_n(y)) < \varepsilon)$.

I can see absolutely no reason why it should, but since I know that equicontinuity and uniform equicontinuity are equivalent when $X$ is compact, I was wondering whether the underlying reason was that when $X$ is compact each continuous $f_n$ is uniformly so.

Henno Brandsma
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James Well
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1 Answers1

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Let $F$ be the family of differentiable functions $f:\mathbb R\to\mathbb R$ such that

(i) $f'$ is bounded

(ii) $|f'(t)|\le |t|$ for all $t$.

(The bound in (i) is allowed to depend on $f$.)

Then (i) shows every $f\in F$ is uniformly continuous, and (ii) shows that $F$ is (pointwise) equicontinuous, but $F$ is not uniformly equicontinuous.

(For example, consider $f_n$ with $f_n'(t)=\min(n, |t|)$: If $s,t>n$ then $|f_n(s)-f_n(t)|=n|s-t|$.)

David C. Ullrich
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