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Let $$\cal A=\left\{ f\in C([0,1]): |f(x)-f(y)|\leq5|x-y| \, \text{ for all } x,y\in [0,1]\right\}$$ What's the closure of $\cal A$ in $C([0,1])$ with respect to the Supremum-Norm?

I think $\cal A$ is already closed and here is my potential proof: The map $C([0,1])\to \overline{ \Bbb R}$ sending $f$ to its Lipschitz constant if it exists, and to $\infty$ if there is no Lipschitz constant is continuous (is that true?) and therefore $\cal A$ is closed since it is the pre-image of the closed set $\{5\}\subset \Bbb R$.

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    $\begingroup$ It would be the pre-image of $[0,5]$, not of $\{5\}$. But the map $f \mapsto \operatorname{Lip}(f)$ is not continuous with respect to the supremum norm. $2^{-n}\cdot \sin (4^nx)$ has norm $2^{-n}$ and Lipschitz constant $2^n$. However, the map is lower semicontinuous. $\endgroup$ – Daniel Fischer Nov 8 '15 at 14:27
  • $\begingroup$ Good points. What's the best way to go about the exercise then? $\endgroup$ – Casey Harrison Nov 8 '15 at 14:30
  • $\begingroup$ Essentially, showing that $\mathcal{A}$ is closed is equivalent to showing that $f \mapsto \operatorname{Lip}(f)$ is lower semicontinuous. You can write $\mathcal{A}$ as an intersection of closed sets, or you can show that a pointwise limit of functions in $\mathcal{A}$ belongs to $\mathcal{A}$. At the end of the day, both ways are doing the same, you just look at it from a different angle. $\endgroup$ – Daniel Fischer Nov 8 '15 at 14:34
  • $\begingroup$ @DanielFischer So if $\|f_n-f\|_\infty\to 0$ for $(f_n)\subset \cal A$ then we also have pointwise convergence $f_n(x)\to f(x)$ and therefore $|f(x)-f(y)|=\lim_n |f_n(x)-f_n(y)|\leq 5 |x-y|$? Is that alright? $\endgroup$ – Casey Harrison Nov 8 '15 at 15:32
  • $\begingroup$ That means $f\in \cal A$ $\endgroup$ – Casey Harrison Nov 8 '15 at 15:32

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