Show this sequence is equicontinuous I'm stuck on an analysis problem to which I've reduced to the following, so some assumptions may be superfluous. 
Let $\{ f_n(x) \} \subset C(X,\mathbb{R}^{\geq0})$ (i.e. $f_n$ is continuous and nonnegative)  where $X$ is a compact subset of $\mathbb{R}^n$.
Suppose further that for all $n,x$, $f_{n+1}(x) \leq f_n(x)$ and that $f_n$ converges pointwise to the continuous function $f$.
Show that the sequence $f_n$ is equicontinuous.
Any help on this problem is appreciated. Thanks.
 A: By Dini's theorem http://en.wikipedia.org/wiki/Dini's_theorem, the convergence is uniform.
The sequence is actually uniformly equicontinuous.
Take $\epsilon>0$.
There exists $N$ such that $\sup_X|f_n-f|\leq \epsilon/3$ for all $n\geq N$.
Now $f$ is uniformly continuous on the compact $X$, so there exists $\delta_0>0$ such that $|f(x)-f(y)|\leq \epsilon/3 $ for all $|x-y|\leq \delta_0$.
For all $n\geq N$ and for $|x-y|\leq \delta_0$, we thus have, by triangular inequality
$$
|f_n(x)-f_n(y)|\leq |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)|
$$
$$
\leq 2\sup_X|f_n-f|+|f(x)-f(y)| \leq 3\frac{\epsilon}{3}=\epsilon.
$$
Finally, the functions $f_1,\ldots,f_{N-1}$ are all uniformly continuous and in finite number, so we can clearly find $\delta_1>0$ such that $|f_n(x)-f_n(y)|\leq \epsilon$ for all $|x-y|\leq \delta_1$ and all $n=1,\ldots,N-1$.
I remains to take $\delta=\min\{\delta_0,\delta_1\}$ to get $|f_n(x)-f_n(y)|\leq \epsilon$ for all $|x-y|\leq \delta$ and all $n$.
So the functions $(f_n)$ are uniformly equicontinuous.
A: Since that $f$ is a continuous function, for $\varepsilon>0$ and $x\in X$ exist $\delta>0$ such that
$$|f(y)-f(x)|<\varepsilon, \ \ \ \mbox{if} \ \ \ |y-x|<\delta.$$
Since that the sequence $(f_n)_n$ converge pointwise,
$$|f_n(x)-f(x)|\longrightarrow0, \ \ \ \ |f_n(y)-f(y)|\longrightarrow0,$$
for $n$ sufficiently huge. Using the monotonicity of the sequence, we obtain
$$
\begin{array}{rcl}
|f_1(x)-f_1(y)| & \leq & |f_1(x)-f(x)|+|f_1(y)-f(y)|+|f(y)-f(x)|\\
                & \leq & |f_n(x)-f(x)|+|f_n(y)-f(y)|+|f(y)-f(x)|\longrightarrow0,
\end{array}
$$
when $n\in\mathbb{N}$ is sufficiently huge and $|y-x|<\delta$. This same argument can be used for $f_2, f_3,...,f_n...$. Then, 
$$|f_n(x)-f_n(y)|<\varepsilon, \ \ \ \ \mbox{if} \ \ \ |y-x|<\delta,$$
for all $n\in\mathbb{N}$, like desired.
A: By continuity of f ; for every $\varepsilon > 0$ (given $x\in $X$)$  , there exist $\Delta>0$ such that
$$|f(y)-f(x)|<\varepsilon/3 \ \ \ \mbox{if}  \ \ \ |y-x|<\Delta.$$
Since  the sequence $(f_n)$ converges pointwise,
$|f_N(x)-f(x)|<\varepsilon/3$     and     $|f_N(y)-f(y)|<\varepsilon/3$
 for  sufficiently large $N$.
And for $n\geq$ $N  $ , we have 
$|f_n(x)-f_n(y)|\leq \ $$|f_N(x)-f(x)|$+$|f_N(y)-f(y)|$+$|f(y)-f(x)|$ $<\epsilon$
Then just take $\delta= min $$\{\delta_1,\delta_2,..,\delta_N,\Delta\}$ where $\delta_i$ satisfies $:|f_i(x)-f_i(y)| <  \varepsilon $ if  $|x-y|<\delta_i                $  .
So , by definition , $f_n$ is equicontinuous.
