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Let $(X,d)$ be a compact metric space and

$$\mathcal L = \{ f\in C(X,\mathbb{R^l}): ||f||_u\le 1, |f(x) - f(y)|\le d(x,y)\quad \forall x,y\in X\}$$.

Then what I claim is that $\mathcal L$ is compact.

Therefore I want to use the Arzelà-Ascoli theorem then I have to prove that $\mathcal L$ is closed,equicontinous and pointwise bounded.

For the last two properties I did the following:

We take an arbitrary $x \in X$ then by hypothesis we have that for every $f \in \mathcal L$ we get $max_x |f(x)| \leq 1$ therefore we have that $$\{f(x): f \in \mathcal L \}\subset B_1(0)$$

Now, for equicontinuity, If we pick an arbitrary $f \in \mathcal L $ and $|x-y|=d(x,y)<\delta=\epsilon$ we get by definition of $ \mathcal L $ that

$$|f(x)-f(y)|<|x-y|=d(x,y)<\delta$$

So Am I right in these issues?, And How can I prove that $\mathcal L$ is closed (I have tried taking a sequence $f_n$ in $\mathcal L $ such that in converges to $f$ and I want to prove that $f \in \mathcal L $ but I don't know how to get that $|f|<1$ and $|f(x)-f(y)|<\epsilon$ with only the triangle inequality)?

Thanks a lot in advance for your help.

user162343
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  • Why was deleted the answer? – user162343 Sep 17 '15 at 16:35
  • I am back in the discussion :) then, what can be done to get closeness? – user162343 Sep 17 '15 at 23:43
  • I am back in the discussion :) then, what can be done to get closeness? – user162343 Sep 18 '15 at 15:30
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    In the argument for pointwise boundedness, you have a slight mistake, if $B_1(0)$ denotes the open ball. $\mathcal{L}$ is defined so that $\lvert f(x)\rvert \leqslant 1$ for all $x$, so a constant function of modulus $1$ belongs to $\mathcal{L}$. Thus all you can say is that for $f\in \mathcal{L}$ and $x\in X$, $f(x)$ belongs to the closed ball with radius $1$ and centre $0$. Or that it belongs to an open ball with radius $r > 1$ [and it belongs to all such balls] and centre $0$. If $B_1(0)$ denotes the closed ball, that part is correct. But later you write that you want to show … – Daniel Fischer Sep 18 '15 at 20:25
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    … $\lvert f\rvert < 1$ for a limit $f$ of a sequence in $\mathcal{L}$, which indicates that it is the open ball. However, what you actually need is $\lvert f\rvert \leqslant 1$ (and you can't get the strict inequality). So, for the limit $f$, you need to show that for arbitrary $x\in X$ you have $\lvert f(x)\rvert \leqslant 1$, and for arbitrary $x,y\in X$ you have $\lvert f(x) - f(y)\rvert \leqslant d(x,y)$. Both follow easily from considering the pointwise limit. – Daniel Fischer Sep 18 '15 at 20:26
  • Well may be you should check this one http://math.stackexchange.com/questions/1441195/proving-closeness-of-a-set-of-continuous-functions/1441221?noredirect=1#comment2936086_1441221 – user162343 Sep 18 '15 at 20:30

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