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A textbook I'm reading states the set $A = \{f \in C^1[0,1]: \|f\|_\infty + \|f'\|_\infty \leq \gamma\}$ is compact in $C[0,1]$ with the sup norm. Here, $\gamma$ is a positive constant.

Well, by Arzela Ascoli, as this set is equicontinuous and uniformly bounded in $C[0,1]$, any sequence in this set has a subsequence which converges uniformly to a function in $C[0,1]$.

What I'm having trouble proving is that this limit is in $A = \{f \in C^1[0,1]: \|f\|_\infty + \|f'\|_\infty \leq \gamma\}$.

I don't see why this limit is necessarily differentiable in the first place. If $f_n \in A$ for all $n$ and $\|f_n - f\|_\infty \rightarrow 0$, then for $f$ to belong to $A$, it would be enough if $\{f_n'\}$ was a Cauchy sequence. (Then $f'$ must necessarily exist and be the limit of this sequence) - but again, I don't think this is true.

Anu
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  • The statement looks false (as you well notice, a uniform limit of $\mathscr{C}^1$ functions has little reason to be $\mathscr{C}^1$ as well – think of a piecewise polynomial approximation of the absolute value). However, if you replace $|f’|_{\infty}$ by the best Lipschitz constant of $f$, then it becomes true. – Aphelli Oct 10 '20 at 09:28
  • @Mindlack Nice observation regarding the Lipschitz constant. This does give us almost everywhere differentiability of the limit function, interestingly. – Anu Oct 10 '20 at 09:42
  • @Caffeine - yes, that counterexample seems to work! I wonder why the textbook says that - at any rate, it used the "compactness" of this set to conclude that if $f$ is an injective and continuous function defined on $A$, then its inverse defined on its range is also continuous. I believe this result still holds, even if $A$ is not compact - it is enough that closed subsets of $A$ are compact, correct me if I'm wrong – Anu Oct 10 '20 at 13:10
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    @Anu It is indeed sufficient, provided $f(A)$ is Hausdorff –  Oct 10 '20 at 13:20
  • @Caffeine Do we need that assumption? Edit: Nevermind, got it. We used it for the implication compact $\Rightarrow$ closed, I think. – Anu Oct 11 '20 at 09:09

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