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Let $X=C([a,b])$ equipped with the norm $\Vert.\Vert_{\infty}$. The closed subspace $F$ of $\alpha$-Holder continuous ($0\lt\alpha\le1)$ functions $F\subset X$ is finite dimensional because if I take a closed unit ball in it, the ball is compact (using Ascoli-Arzelà). What can we say about Lipschitz continuous functions space? Can we argue this way and conclude that it is finite dimensional?

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  • $\begingroup$ I think the same proof should work. $\endgroup$ – pitariver Jun 26 at 13:19
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Your argument in the case $\alpha \in (0,1)$ is not true. Arzela-Ascoli asserts that the ball $$ \{ f | \|f\|_{0,\alpha} \le 1\}, $$ where $\|\cdot\|_{0,\alpha}$ denotes the Hölder norm, is compact in the space $(C([a,b]), \|\cdot\|_\infty)$. It does not implies compactness in the Hölder space $(C^{0,\alpha}([a,b]), \|\cdot\|_{0,\alpha})$!

In fact, $C^{0,\alpha}([a,b])$ is infinite dimensional and, thus, cannot have a compact unit ball.

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No. $\{x\mapsto \lvert x-y\rvert: y\in(a,b)\}$ are linearly independent 1-Lipschitz functions in $C([a,b])$.

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  • $\begingroup$ Why the proof doesn't work with α=1? technically for α in that range Lipschitz functions are a subspace of α-Holder continuous functions $\endgroup$ – banach-alaoglu-zielony Jun 26 at 13:49
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$1,x,x^{2}/2,x^{3}/3,...$ are linearly independent elements of this space when $a=0$ and $b=1$.

For equicontinuity you need a condition like $|f(x)-f(y)| \leq M|x-y|$ for all $f$ in $F$ with $M$ independent of $f$. Since such a condition is not available your argument fails. For no value of $\alpha$ is your space finite dimensional.

PS: linear independence of the sequence I have defined is obvious since no non-zero polynomial can have infinitely many zeros.

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  • $\begingroup$ Why the proof doesn't work with $\alpha=1$? technically for $\alpha$ in that range Lipschitz functions are a subspace of $\alpha$-Holder continuous functions $\endgroup$ – banach-alaoglu-zielony Jun 26 at 13:43
  • $\begingroup$ Equicontinuity is not true. You are considering functions with sup norm bounded by 1 but the constant in Lipschitz condition is not bounded. $\endgroup$ – Kavi Rama Murthy Jun 26 at 13:56

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