# Is $F=\{f\in C^1([0,1]):~ |f(x)|\leq 1 ~\text{and}~ 1\leq f'(x)\leq 2 ~\text{for all }x\in [0,1]\}$ compact?

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?

• why is $F_x$ included in $[0,1]$? – supinf Nov 22 '19 at 12:19
• Oh mistype, changed. – Evgeny Kuznetsov Nov 22 '19 at 12:36

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 $$F$$.