Determine if the family, $\mathcal{F}$ = sequence of $f_{n}(x)$ that are $C_{n}$ - Lipschitz and $C_{n} \to C$ in $[0,1]$ is compact.
To determine if this set of functions is compact we have to establish the set is:
- bounded
- closed
- equicontinuous
Proof:
i) Equicontinuity:
It is known that all Lipschitz functions are continuous. In this case let $\delta = \frac{\epsilon}{C_{n}}$:
$$\therefore |f_{n}(x) - f_{n}(y)| \leq C_{n}|x-y| < C_{n} \frac{\epsilon}{C_{n}} = \epsilon$$
So $\mathcal{F}$ is actually uniformly equicontinuous.
ii) Bounded: (I am having some troubles with this and the next)
Want to show: $\exists M > 0$ such that $|f_{n}| \leq M$, for all $f_{n} \in \mathcal{F}$ on $[0,1]$
Since these functions are Lipschitz:
$$ |f_{n}(x) - f_{n}(y)| \leq C_{n}|x-y|$$
Now I was thinking since I am on $[0,1]$ I could perhpas do something like let $x = 0, y = 1$. Then that would mean:
$$ |f_{n}(x) - f_{n}(y)| \leq C_{n}$$
I feel that I should be able to get some sort of bound from this, but perhaps I am writing something wrong in terms of my interpretation.
iii) Closed
Want to show: $f \in \mathcal{F}$, which is the same as $|f(x) - f(y)| \leq C|x-y|$
Suppose that $f_{n}(x)$ converges uniformly to $f$. This means that for all $\epsilon > 0$, there exists a $N > 0$ such that $|f_{n}(x) - f(x)| < \frac{\epsilon}{3}$, for all $n \geq N$ and $x,y \in [0,1]$.
Putting this together: $$|f(x) - f(y)| \leq |f(x) - f_{n}(x)| + |f_{n}(x) - f_{n}(y)| + |f_{n}(y) - f(y)| < \frac{\epsilon}{3} + C_{n}|x-y| + \frac{\epsilon}{3} $$
Again I am having som eissues with the final statement. I envisioned something of the sort $C_{n}|x-y|$ appearing in all of my terms and then taking $\lim_{n \to \infty}$, but it didn't work out how I had hoped.