Proving $f$ is Lipschitz given $f_{n}\rightarrow f$ and $f_{n}$ is uniformly Lipschitz

Question

Given $f_{n} : [a,b]\rightarrow \mathbb R$ is uniformly Lipschitz and $f_{n}\rightarrow f$, how can I show that $f$ is Lipschitz?

$f_{n}$ is uniformly Lipschitz $\leftrightarrow \exists K > 0$ such that for all $n$ and for every $x,y\in [a,b]$ we have $|f_{n}(x)-f_{n}(y)| < K|x-y|$.

$f_{n}\rightarrow f \leftrightarrow \forall~\epsilon>0~~\exists~N$ such that $\forall~n>N$ we have $\mid f_{n}(x) - f(x)\mid<\epsilon~~~(=\mid x-y\mid?)$

Then we have

\begin{align} |f(x)-f(y)| &\leq |f(x)- f_{n}(x)| + |f_{n}(y)-f(y)| + |f_{n}(x)-f_{n}(y)| \\ &\leq |x-y| + |x-y| + K|x-y| \\ &\leq (K + 2) |x - y| \end{align}

which shows that $f$ is Lipschitz. Can anybody tell me if this makes sense? Sorry if layout is poor.

Answer 1

Ypu are almost correct.

Ιn your proof you must take $n_1$ for $x$ and $n_2$ for $y$ and then take $N=\max\{n_1,n_2\}$

After that the inequalities are correct.

There is a simpler proof:

If $x,y \in [a,b]$ then $$|f(y)-f(x)|=\lim_n|f_n(x)-f_n(y)| \leq K|x-y|$$

Answer 2

You have $$|f(x) - f(y)| \le 2 \varepsilon + K|x - y|.$$

Then take $\epsilon \to 0$ to obtain $$|f(x) - f(y)| \le K|x - y|.$$

But as Marios said: you need an $N$ for $x$ and an $N$ for $y$.

There is no relation between $\varepsilon$ and $|x - y|$ except that as $\varepsilon \to 0$, eventually $\varepsilon < |x - y|$.

Answer 3

Start with the uniform Lipschitz inequality ($x,y$ fixed but arbitrary) $$ |f_n(x)-f_n(y)|\le K|x-y| $$ and use the fact that inequality is preserve in the limit to let $n\to\infty$. There results $$ |f(x)-f(y)|\le K|x-y|, $$ and $f$ is seen to be Lipschitz.