$g_{n}(x):=max \lbrace f_{1}(x),....\rbrace$ uniformly convergent if $\lbrace f_{n} \rbrace_{n \in \mathbb{N}}$ uniformly bounded and equicontinuous Let $\lbrace f_{n} \rbrace_{n \in  \mathbb{N}}$ a sequence of functions with real values which is uniformly bounded and equicontinuous in a metric space $X$. For each $n \in \mathbb{N}$, we define the following  functions $g_{n}:X \to \mathbb{R}$:
$$g_{n}(x):=max \lbrace f_{1}(x), f_{2}(x),....,f_{n}(x)\rbrace$$
For each $x \in \mathbb{R}$. Prove that the sequence $\lbrace g_{n} \rbrace_{n \in  \mathbb{N}}$ is uniformly convergent.
Since $\lbrace f_{n} \rbrace_{n \in  \mathbb{N}}$ is uniformly bouded there is an $M$ such that $|f_{n}(x)| \leq M$ for every $n \in \mathbb{N}$ and $x \in X$. Also, as this sequence is also equicontinuous on $X$ we have that for every $\epsilon >0$ there is a $\delta>0$ such that if for every $x,y \in X$ and $d_{X}(x,y) < \delta$, then $d_{\mathbb{R}}(f_{n}(x), f_{n}(y)) < \epsilon$ for every $n \in \mathbb{N}$.
First and foremost, we need to find the pointwise limit of the sequence $\lbrace g_{n} \rbrace_{n \in  \mathbb{N}}$. Its not difficult to see this an incresing sequences, so that  $f_{n} \leq f_{n+1}$ for every $n \in \mathbb{N}$. On the other hand, since $\lbrace f_{n} \rbrace_{n \in  \mathbb{N}}$ is uniformly bounded this sequence of functions is bounded. So by some well known sequence theorem we got that $\lbrace g_{n} \rbrace_{n \in  \mathbb{N}} \to sup (\lbrace f_{n} \rbrace_{n \in  \mathbb{N}})=s$ right? Still Im not sure how to prove than $\lbrace g_{n} \rbrace_{n \in  \mathbb{N}}$ uniformly converges to $s$ :/
 A: It seems that if $X$ is not compact, the proposition is false. Consider
the following example:
Let $X=[0,\infty)$ equipped with the usual metric. For each $n$,
let $f_{n}:X\rightarrow[0,\infty)$ be defined by
$$
f_{n}(x)=\begin{cases}
1, & \mbox{if }x\in[0,n]\\
\mbox{linear }, & \mbox{if }x\in(n,n+1)\\
0 & \mbox{if }x\in[n+1,\infty)
\end{cases}.
$$
Clearly $\{f_{n}\mid f\in\mathbb{N}\}$ is uniformly bounded and uniformly equicontinuous.
Moreover, $f_{1}\leq f_{2}\leq\ldots$. Therefore $g_{n}=\max\{f_{1},\ldots,f_{n}\}=f_{n}$.
Clearly $g_{n}\rightarrow1$ pointwisely but the not uniformly.
A: Lemma 1: Let $(X,d)$ be a compact metric space. Let $f_{n}:X\rightarrow\mathbb{R}$.
If $\{f_{n}\mid n\in\mathbb{N}\}$ is equicontinuous, then it is uniformly
equicontinuous.
Proof of Lemma 1: Let $\varepsilon>0$ be given. For each $x\in X$,
there exists $\delta_{x}>0$ such that $|f_{n}(y)-f_{n}(x)|<\varepsilon$
whenever $y\in B(x,\delta_{x})$ and $n\in\mathbb{N}$. Note that
$\{B(x,\delta_{x}/2)\mid x\in X\}$ is an open covering of the compact
set $X$, so it admits a finite subcover $\{B(x_{i},\delta_{x_{i}}/2)\mid i=1,\ldots,m\}$.
Let $\delta=\min_{1\leq i\leq m}\delta_{x_{i}}>0.$ Let $y_{1,}y_{2}\in X$
be arbitrary such that $d(y_{1},y_{2})<\delta$. Let $n\in\mathbb{N}$
be arbitrary. Choose $i$ such that $y_{1}\in B(x_{i},\delta_{i})$,
then
\begin{eqnarray*}
d(y_{2},x_{i}) & \leq & d(y_{2},y_{1})+d(y_{1},x_{i})\\
 & < & \delta+\frac{1}{2}\delta_{x_{i}}\\
 & \leq & \delta_{x_{i}}.
\end{eqnarray*}
Clearly, we also have $d(y_{1},x_{i})<\delta_{x_{i}}.$ Therefore,
$|f_{n}(y_{1})-f_{n}(y_{2})|\leq|f_{n}(y_{1})-f_{n}(x_{i})|+|f_{n}(x_{i})-f_{n}(y_{2})|<2\varepsilon.$
Hence, $\{f_{n}\mid n\in\mathbb{N}\}$ is uniformly equicontinuous.

Lemma 2: Let $(X,d)$ be a metric space. Let $f_{n}:X\rightarrow\mathbb{R}$.
Suppose that $\{f_{n}\mid n\in\mathbb{N}\}$ is equicontinuous. For
each $n\in\mathbb{N}$, define $g_{n}:X\rightarrow\mathbb{R}$ by
$g_{n}=\max\{f_{1},f_{2},\ldots,f_{n}\}$. Then $\{g_{n}\mid n\in\mathbb{N}\}$
is equicontinuous.
Proof of Lemma 2: Let $\varepsilon>0$ and $x\in X$ be given. Choose
$\delta>0$ such that $|f_{n}(x)-f_{n}(y)|<\varepsilon$ whenever
$y\in B(x,\delta)$ and $n\in\mathbb{N}$. Let $y\in B(x,\delta)$
and $n\in\mathbb{N}$ be arbitrary. Choose $k\in\{1,\ldots,n\}$ such
that $g_{n}(x)=f_{k}(x)$. For any $j=1,\ldots,n$,
\begin{eqnarray*}
f_{j}(y) & < & f_{j}(x)+\varepsilon\\
 & \leq & g_{n}(x)+\varepsilon.
\end{eqnarray*}
Therefore $g_{n}(y)=\max_{1\leq j\leq n}f_{j}(y)<g_{n}(x)+\varepsilon$.
On the other hand,
\begin{eqnarray*}
g_{n}(y) & \geq & f_{k}(y)\\
 & > & f_{k}(x)-\varepsilon\\
 & = & g_{n}(x)-\varepsilon.
\end{eqnarray*}
This shows that $|g_{n}(y)-g_{n}(x)|<\varepsilon$. Hence, $\{g_{n}\mid n\in\mathbb{N}\}$
is equicontinuous.

Now, we go back to your question and prove the following:
Proposition: Let $(X,d)$ be a compact metric space. For each $n\in\mathbb{N}$,
let $f_{n}:X\rightarrow\mathbb{R}$ be a function. Define $g_{n}:X\rightarrow\mathbb{R}$
by $g_{n}=\max\{f_{1},f_{2},\ldots,f_{n}\}$. Suppose that $\{f_{n}\mid n\in\mathbb{N}\}$
is uniformly bounded and equicontinuous, then the sequence of functions
$(g_{n})$ converges uniformly.
Proof of Proposition: Choose $M>0$ such that $|f_{n}(x)|\leq M$
for all $x\in X$ and $n\in\mathbb{N}$. Note that for each $x\in X$,
$(g_{n}(x))_{n}$ is bounded and monotonic increasing, so $\lim_{n\rightarrow\infty}g_{n}(x)$
exists. Define $g:X\rightarrow\mathbb{R}$ by $g(x)=\lim_{n\rightarrow\infty}g_{n}(x)$.
Firstly, we go to show that $g$ is continuous (actually uniformly continuous. However, we do not need this fact at this moment). By Lemma 1 and Lemma
2, $\{g_{n}\mid n\in\mathbb{N}\}$ is uniformly equicontinuous. Let
$x\in X$ and $\varepsilon>0$ be given. Choose $\delta>0$ such that
$|g_{n}(s)-g_{n}(t)|<\varepsilon$ whenever $d(s,t)<\delta$ and $n\in\mathbb{N}$.
Let $y\in B(x,\delta)$ be arbitrary, then for each $n$, $|g_{n}(y)-g_{n}(x)|<\varepsilon$.
Letting $n\rightarrow\infty$, we obtain $|g(y)-(x)|\leq\varepsilon$.
This shows that $g$ is continuous at an arbitrary point $x$.

We go to prove that $g_{n}\rightarrow g$ uniformly by contradiction.
Suppose the contrary that $(g_{n})$ does not converges to $g$ uniformly,
then may choose $\varepsilon>0$ such that
$$
\forall N\exists n\geq N\exists x\in X\left(|g_{n}(x)-g(x)|\geq\varepsilon\right).
$$
Put $N=1$, we obtain $n_{1}\geq1$ and $x_{1}\in X$ such that $|g_{n_{1}}(x_{1})-g(x_{1})|\geq\varepsilon$.
Put $N=n_{1}+1$, we obtain $n_{2}\geq n_{1}+1$ and $x_{2}\in X$
such that $|g_{n_{2}}(x_{2})-g(x_{2})|\geq\varepsilon$. Continue
the process indefinitely, we obtain sequences $(n_{k})_{k}$, $(x_{k})$
such that $n_{1}<n_{2}<\ldots$ and $|g_{n_{k}}(x_{k})-g(x_{k})|\geq\varepsilon$
for all $k$. (If we want to argue it formally, we need to invoke
the Recursion Theorem). Since $X$ is compact, we may choose a subsequence
$(x_{k_{l}})$ of $(x_{k})$ such that $x_{k_{l}}\rightarrow x$ for
some $x\in X$. To simplify notation, we denote $h_{l}=g_{n_{k_{l}}}$
and $y_{l}=x_{k_{l}}$, for $l\in\mathbb{N}$. Hence, $|h_{l}(y_{l})-g(y_{l})|\geq\varepsilon$
for all $l$ and $y_{l}\rightarrow x$.
On the other hand, note that $\{g_{n}\mid n\in\mathbb{N}\}$ is equicontinuous
(by Lemma 2), which implies that the sub-family $\{h_{l}\mid l\in\mathbb{N}\}$
is also equicontinuous. By Lemma 1, $\{h_{l}\mid l\in\mathbb{N}\}$
is uniformly equicontinuous. Therefore, there exists $\delta_{1}>0$
such that $|h_{l}(s)-h_{l}(t)|<\varepsilon/10$ whenever $d(s,t)<\delta_{1}$
and $l\in\mathbb{N}$. Since $g$ is continous at $x$, there exists
$\delta_{2}>0$ such that $|g(t)-g(x)|<\varepsilon/10$ whenever $t\in B(x,\delta_{2})$.
Since $y_{l}\rightarrow x$, we can choose $L_{1}\in\mathbb{N}$ such
that $d(y_{l},x)<\min(\delta_{1},\delta_{2})$ whenever $l\geq L_{1}$.
Note that $h_{l}(x)\rightarrow g(x)$, so there exists $L_{2}$ such
that $|h_{l}(x)-g(x)|<\varepsilon/10$ whenever $l\geq L_{2}$. Put
$l=\max(L_{1},L_{2})$, then
\begin{eqnarray*}
\varepsilon & \leq & \left|h_{l}(y_{l})-g(y_{l})\right|\\
 & \le & \left|h_{l}(y_{l})-h_{l}(x)\right|+\left|h_{l}(x)-g(x)\right|+\left|g(x)-g(y_{l})\right|\\
 & < & \varepsilon/10+\varepsilon/10+\varepsilon/10\\
 & = & \frac{3\varepsilon}{10},
\end{eqnarray*}
which is a contradiction.
In the above, have used the following facts:
(1) $l\geq L_{1}\Rightarrow d(y_{l},x)<\min(\delta_{1},\delta_{2})\leq\delta_{1}\Rightarrow|h_{l}(s)-h_{l}(t)|<\varepsilon/10$,
(2) $l\geq L_{2}\Rightarrow|h_{l}(x)-g(x)|<\varepsilon/10$,
(3) $l\geq L_{1}\Rightarrow d(y_{l},x)<\min(\delta_{1},\delta_{2})\leq\delta_{2}\Rightarrow\left|g(x)-g(y_{l})\right|<\varepsilon/10$.
