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Inspired by the old question I considered the following

Let $$F=\{f\in C^1([0,1]):~~ |f(x)|\leq 1 ~~\text{and}~~ 1\leq f'(x)\leq 2 ~~\text{for all }x\in [0,1]\}$$ the set of continuous, bounded by $-1$ and $1$, functions $f:[0,1]\to \mathbb R$ with bounded, by $1$ and $2$, continuous first derivative.

In this case I can not apply the trick, which is applied there, and I guess $F$ is compact in this case, but I can not prove it yet.

Is my guess right or rather $F$ is only pre-compact (subset closure of which is compact)?

Theorem(Arzela-Ascoli):

Let $X$ be a compact metric space, $(Y,d)$ any metric space and $F$ a subset of $C(X,Y)$. Then $\operatorname{cl}(F)$ is compact iff the following two conditions are valid:

  • $F$ is equicontinuous
  • for each $x\in X$, the subset $F_x=\operatorname{cl}\left(\{f(x):~f\in F\}\right)$ is a compact subspace of $Y$.

where $\operatorname{cl}(A)$ is a closure of $A$.

Is $F$ closed subset of $C([0,1])$?

If $F$ is closed one can apply Arzela-Ascoli theorem to show that $F$ is compact, since boundedness of $F_x$ and equicontinuity of $F$ are clear in this case.

Lemma: If $F$ is an equicontinuous subspace of $C(X,Y)$ then so is $\operatorname{cl}(F)$.

By this lemma $F_x=\operatorname{cl}\left(\{f(x):~ f\in F\}\right)=\operatorname{cl}\left(\{f(x):~ f\in \operatorname{cl}(F)\}\right)$, am I right?

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  • $\begingroup$ why is $F_x$ included in $[0,1]$? $\endgroup$ – supinf Nov 22 '19 at 12:19
  • $\begingroup$ Oh mistype, changed. $\endgroup$ – Evgeny Kuznetsov Nov 22 '19 at 12:36
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Hint: It is possible to show that $$ F_x = [x-1,x] $$ explicitly. To gain intuition of this fact it might help to think about what functions are in $F$ what and functions are not in $F$.

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