Given a sequence of functions $(f_n)$, with $f_n : [a, b] → \mathbb{R}$ we say that it is uniformly Lipschitz if there exists $K > 0$ such that for every $n$ and every $x, y ∈ [a, b]$ we have
$|f_n(x) − f_n(y)| ≤ K|x − y|$
Prove that if $f_n → f$ and (f_n) is uniformly Lipschitz then $f$ is Lipschitz (i.e. there exists $M > 0$ such that for every $x, y ∈ [a, b]$ we have $|f(x) − f(y)| ≤ M|x − y|)$.
I have tried the triangle inequality, as in adding $|f_n(x) − f(x)|$, $|f(y) − f_n(y)|$, $|f(x) − f(y)|$ is greater than $|f_n(x) − f_n(y)|$, does this help?
$$|f_n(x)-f_n(y)|\leq K|x-y|,\tag{1}$$ for all $n$. Since $f_n(x)\to f(x)$ and $f_n(y)\to f(y)$, taking the limit $n\to \infty $ in $(1)$ yields $$|f(x)-f(y)|\leq K|x-y|.$$