Prove that $M_\epsilon$ is compact for any $\epsilon>0.$ A continuous functions $ f(x), x \in [a,b]$ is called convex if for any 
$ x_1, x_2 \in [a,b]$, and for any $\lambda \in [0,1]$. 
$ f((1-\lambda )x_1 + \lambda x_2 ) \le ( 1- \lambda) f(x_1)+ \lambda f(x_2) $.
Let $M_\epsilon \subset C[0,1]$ be the set of functions such that if $ f \in M_\epsilon$ then there exists a function $ g \in C[-\epsilon,1+\epsilon ] $ Such that: (1) $ |g(x)| \le 1 $ in $C[-\epsilon,1+\epsilon ] $;
(2) $g(x)$ is convex on $C[-\epsilon,1+\epsilon ] $; and
(3) $g(x)=f(x)$ for any $x \in [0,1] $.
Prove that $M_\epsilon$ is compact for any $\epsilon>0.$
Proof : Let the set $M_\epsilon$ is relatively compact, by the Arzela-Ascoli theorem. 
Now we can apply it here because, if $f\in M_\epsilon$ :


*

*If $x\in[0,1]$, then $|f(x)|=|f(x)-f(0)|\le M/2|x-0|\le M/2;$

*If $ x_1,x_2\in[0,1], if \epsilon>0 $ and if we take,
$\delta=(2\epsilon)/M,$ 
Then:
$|x-y|<\delta$
$\Rightarrow|f(x)-g(x)|<\epsilon$
The set $M_\epsilon$ is also closed.
Therefore, it is compact.
Does the proof is good ?
 A: Assume that $f_n\in M_\varepsilon$ and $f_n\rightarrow f$. 
Here it is uniform convergence so that $f$ is continuous and
convex.
If $F_n$ is extension of $f_n$ on
$[-\varepsilon,1+\varepsilon]$,
then there is subsequence $F_n$ s.t. $
F_n(-\varepsilon)\rightarrow a,\ F_n(1+\varepsilon)\rightarrow b$.
Define $F$ to be extension of $f$ s.t. $F(-\varepsilon)=a,\ F(1+
\varepsilon)=b$ and $F|[-\varepsilon,0],\ F|[1,1+\varepsilon]$ are linear functions.
Here $G_n(x)=\sup_x\ \{F(x),F_n(x)\}$ is convex and note that
$G_n\rightarrow F$. Hence $F$ is convex.
A: Assume that $ f_n∈M_ε $ and $f_n→f$.
Here it is uniform convergence so that f is continuous and convex.
If $F_n$ is extension of $f_n$ on [−ε,1+ε], then there is subsequence $F_n$ s.t. $F_n(−ε)→a, F_n(1+ε)→b.$
Define F to be extension of ff s.t. F(−ε)=a, F(1+ε)=b and F|[−ε,0], F|[1,1+ε] are linear functions.
Here $G_n(x)=sup_x {F(x),F_n(x)}$ is convex and note that $G_n→F.$ Hence F is convex.
Let the set $M_\epsilon$ is relatively compact, by the Arzela-Ascoli theorem. Now we can apply it here because, if $f\in M_\epsilon$ :\
 * If $x\in[0,1]$, then $|f(x)|=|f(x)-f(0)|\le M/2|x-0|\le M/2;$\
 * If $ x_1,x_2\in[0,1], if \epsilon>0 $ and if we take $\delta=(2\epsilon)/M,$ \
 Then:
 $|x-y|<\delta$\
 $\Rightarrow|f(x)-g(x)|<\epsilon$\
 The set $M_\epsilon$ is also closed.\ 
 Therefore, it is closed.
CAN I USE THIS AS A COMPLETE PROOF ?
