Sequence of M-Lipschitz functions

Question

Let $\alpha$ , $\beta \in \mathbb R$ with $\alpha < \beta$, $M \geq 0$ and $(f_n)$ a sequence of M-Lipschitz functions of [$\alpha , \beta$] in $\mathbb R$. If $(f_n)$ converges simply (or "pointwise" in English I think) to a function $f$ on [$\alpha , \beta$], show that $f$ is also M-lipschitz.

We have that $(f_n)$ converges simply so $\forall x \in [\alpha , \beta], \forall \epsilon > 0, \exists n_0$ such that $\forall n > n_0, \lvert f_n(x) - f(x) \rvert \leq \epsilon$

And we also have that the $f_n$ are M-Lipschitz so $\forall x,y \in [\alpha , \beta], \lvert f_n(x) - f_n(y) \rvert \leq M\lvert x-y \rvert$

We want to show that $\forall x,y \in [\alpha , \beta], \lvert f(x) - f(y) \rvert \leq M\lvert x-y \rvert$

I tried to write $\lvert f(x) - f(y) \rvert = \lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert$

Then we have $\lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert \leq \epsilon + M\lvert x-y \rvert + \epsilon$ but it is not really the form we wanted and I think I made some mistakes with the quantifiers. Can someone help me ? (The correction solves this exercise just by saying : when going to the limit the inequality stays true. I tried to find something more precise and clear).

Answer 1

If $x \leq M|x-y| + \epsilon$ for every $\epsilon$, then $x \leq M|x-y|$.

Proof:

Suppose $x > M|x-y|$.

Then $x> M|x-y| - \epsilon$ for every $\epsilon > 0$. Hence $M|x-y| - \epsilon < x\leq M|x-y| + \epsilon \implies$

$-\epsilon <x-M|x-y| \leq \epsilon \implies x-M|x-y|=0$ (which is proven using a basic contradiction).

Hence $x = M|x-y| \implies x > x$, a contradiction.

Your problem uses $2\epsilon$ which is a special case of this.

Answer 2

If $f_n$ is $M$-Lipschitz (with $M$ that does not depend on $n$) then for every $x,y$ in your domain you have $$ |f_n(x)-f_n(y)|\leqslant M|x-y|\quad \forall n\in\mathbb{N}. $$ But the absolute value is a continuous function, as well as the sum of real numbers. Therefore, if $f_n$ converges pointwise to $f$ you can pass to the limit for $n\to\infty$ in the previous relation to get $$ |f(x)-f(y)|\leqslant M|x-y|. $$