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I am solving problems in topology of a previous year paper of our Institute and I could not think of this problem. Please help. Our faculty is unfortunately not interested in teaching. Question is: $A$ is the closure in $C[0, 1]$ of the set $B = \{ f \in C^1([0, 1]) : \rvert f(x) \rvert \le 1 \text{ and } \lvert f'(x) \rvert \le 1 \text{ for all } x \in[ 0,1] \}$. Prove that the set $A$ is connected.

Paul Frost
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1 Answers1

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$A$ is defined as the closure of some set $A' = C([0,1],B)$, probably in the space $C([0,1], C^1[0,1])$ if I understood your question correctly. To show that $A$ is connected, it is enough to show that $A'$ is connected, because the closure of a connected set is connected.

The easiest way to show that $A'$ is connected is probably to show that $A'$ is convex: For any $F, G \in A'$, and any $t \in [0,1]$, the map $x \mapsto (1-t)F(x)+tG(x)$ is also in $A'$. Why? Because $B$ is a convex subset of the topological vector space $C^1([0,1])$. I'll leave proving the convexity of $B$ as exercise to you.

Magma
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  • @ Magma are path connectedness and convexity same, just two different terminology? –  Aug 02 '19 at 16:00
  • Path connected means any two points are connected by a path. Convex means any two points are connected by a straight line. Convexity is a much stronger property. – Magma Aug 02 '19 at 17:01
  • @ Magma thank you very much –  Aug 03 '19 at 11:21