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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 ?

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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.

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  • $\begingroup$ Thank you. Should I add this with my proof to complete ? $\endgroup$ – Tazim Taz Dec 11 '17 at 13:23
  • $\begingroup$ Yes. it is no problem. $\endgroup$ – HK Lee Dec 11 '17 at 13:24
  • $\begingroup$ HK LEE : MAY I USE THE PROOF BELOW AS A COMPLETE PROOF OF MY PROBLEM.? $\endgroup$ – Tazim Taz Dec 19 '17 at 19:33
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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 ?

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