Let $M$ be a Metric compact space. $\{f_n\}_n \subseteq \{f:M \to \mathbb{R}| f$ is continuous$\}$. Suppose that:

  • $\lim_{n\to \infty} f_n(x) = 0 \; \forall x \in M$
  • $f_1(x)\geq f_2(x)\geq ...\geq f_n(x)\geq ...,\;\forall x \in M$.

Show that $\{f_n\}_n$ is equi-continuous, that is, $\forall \epsilon>0$, $\forall x \in M$ $\exists\; \delta(\epsilon,x)$ s.t. if $|x-y|<\delta(\epsilon,x) \implies|f_k(x)-f_k(y)|\leq \epsilon \;\forall k$.

I believe I've shown this result without the necessity that $f_i(x)\geq f_{i+1}(x)\;\forall x$ which is concerning because it's unlikely that my professor made this restriction for nothing. So I need to find my mistake.

Lets fix a value of $x \in M$. Let $\epsilon>0$.

As $\lim_kf_k(x) = 0$, exists a $k_0$ such that $|f_k(x)|\leq \frac{\epsilon}{2} \; \forall k \geq k_0$. So $|f_k(x)-f_k(y)|\leq |f_k(x)|+|f_k(y)|\leq \epsilon \forall x,y \in M$ if $k \geq k_0$.

Let us look at $k < k_0$ then.

Every $f_n$ is continuous, therefore $\exists \delta_n(\epsilon,x)$ such that if $|x-y|\leq \delta_n(\epsilon,x) \implies |f_n(x)-f_n(y)|\leq \epsilon$ for $n\in \{1,2,3,...,k_0-1\}$. Take $\delta = min_n\{\delta_n(\epsilon,x)\}$.

Then if $|x-y|\leq \delta \implies |f_k(x)-f_k(y)|\leq \epsilon\; \forall k$

So $\{f_n\}_n$ is equi-continuous.

Where is my mistake ????

  • 3
    $\begingroup$ Your $k_0$ depends on $x$ so you cannot say $\forall x,y$. $\endgroup$ Sep 14, 2018 at 22:09
  • $\begingroup$ Thank you, I found the correct solution here, math.stackexchange.com/q/2913144 I needed to use the Dini's theorem first, before carrying on with my idea. $\endgroup$ Sep 15, 2018 at 20:30

1 Answer 1


You fixed $x.$ Now for any $\epsilon >0$ and for any fixed $y$ there exists $K(x,y,\epsilon)$ such that $n>K(x,y,\epsilon)\implies |f_n(x)|+|f_n(y)|<\epsilon.$

It does not follow from this that there exists $K'(\epsilon)$ such that $\forall x',y'\in M \;(|f_n(x')|+|f_n(y')|<\epsilon).$


It is necessary that $(f_n(x))_{n\in \Bbb N}$ is decreasing for each $n.$ For example: Let $M=[0,1].$ Let $f_n(x)=n^2x$ for $x\in [0,1/2n]$ and $f_n(x)=n-n^2x$ for $x\in [1/2n,1/n]$ and $f_n(x)=0$ for $x\in [1/n,1].$ Then $f_n\to 0 $ pointwise but $\{f_n:n\in \Bbb N\}$ is not equicontinuous.

It is necessary that $M$ is compact. For example: Let $M=[0,\infty)$ and let $f_n(x)=x/n$. Then every sequence $(f_n(x))_{n\in \Bbb N}$ is decreasing to $0,$ but $\{f_n:n\in \Bbb N\}$ is not equicontinuous.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.