Proving closeness of a set of continuous functions

Question

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

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

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, and I think I should use compactness but I don't how uniformly continuity helps here)?

Thanks a lot in advance for your help.

For the second one I've got

$$|f(x)-f(y)|=|f(x)-f_n(x)+f_n(x)-f(y)| \le |f(x)-f_n(x)|+|f_n(x)-f(y)|=|f(x)-f_n(x)|+|f_n(x)-f_n(y)+f_n(y)-f(y)|\le \frac{\epsilon}{2}+d(x,y)+\frac{\epsilon}{2}=d(x,y)+\epsilon$$

And we are done right?

Hint: $f_n\in\mathcal L$ such that $f_n\to f\in C(X,\mathbb R)$ uniformly. What is $| f |_u$? Use $f = f - f_n + f_n$. — user251257, Sep 18 '15 at 15:48
How do you use the last issue ?because you will get $||f||_u=||f-f_n+f_n||_u \le ||f-f_n||_u+||f_n||_u \le \epsilon +1$ — user162343, Sep 18 '15 at 16:05
As small as we want right :), and for the other condition? Thnks — user162343, Sep 18 '15 at 16:08
Nowhere. You could do the same thing on space of bounded functions. I suspect the next assignment is to show that $\mathcal L$ is compact — user251257, Sep 18 '15 at 16:13
Right :), but by Arzelá-Ascoli we only have to prove pointwise bounded, and equicontinuity where we use the unit ball since $||f||_u \le 1$ and $\delta = \epsilon $ :) Am I right? — user162343, Sep 18 '15 at 16:16
Ok :) That is true :), Now I am sorry for this but can you help me with this one http://math.stackexchange.com/questions/1438366/how-to-use-arzel%C3%A0-ascoli-theorem-in-this-situation please? Thanks in advance — user162343, Sep 18 '15 at 16:22

score 1 · Answer 1 · answered Sep 18 '15 at 15:52

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I assume you want to show that the set is closed in $C$. Quite obviously the set $||f||\le 1$ is closed, so it suffices to show that if $f_n\rightarrow f$ in the norm with $f_n$ in $L$ $\Rightarrow f\in L$. For this it is sufficient that $|f(x)-f(y)|\le d(x,y)$ which follows by uniform convergence (which implies pointwise convergence for every fixed pair $(x,y)$).

answered Sep 18 '15 at 15:52

Thomas

22,361

Thanks a lot @Thomas but Can you explain me more about your answer is because I didn't get it :) Thanks – user162343 Sep 18 '15 at 16:03
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@user162343 Which part did you not get? Is it clear to you that the ball under consideration of radius 1 is closed? If yes, then you are looking at the intersection of a closed set ($||f||\le1)$ and another set (the one with the condition on $d$) Since the intersection of two closed sets is closed, you only need to show that the second set is closed. Since you are in a normed space you only need to show that a sequence in that set which converges in the ambient space has the limit in said set. That follows from what I wrote. – Thomas Sep 18 '15 at 18:07

Proving closeness of a set of continuous functions

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