# Let $\{f_n\}_n$ continuous functions such that $\lim f_n(x) = 0 \; \forall x$. Show that they are equicontinuous

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 ????

• Your $k_0$ depends on $x$ so you cannot say $\forall x,y$. Sep 14, 2018 at 22:09
• 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. Sep 15, 2018 at 20:30

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.