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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 ????

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    $\begingroup$ Your $k_0$ depends on $x$ so you cannot say $\forall x,y$. $\endgroup$ – Eclipse Sun Sep 14 '18 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. $\endgroup$ – Matheus barros castro Sep 15 '18 at 20:30
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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 $\forall x',y'\in M \;(|f_n(x')|+|f_n(y')|<\epsilon).$

Remarks:

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.

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