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Let $f_n: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable for each $n \in \mathbb{N}$ with $|f'_n(x)| \le 1$ for all $n$ and $x$. Let $\lim_{n \rightarrow \infty} f_n(x) = g(x)$ for all $x$. Prove that $g: \mathbb{R} \rightarrow \mathbb{R}$ is Lipschitz-continuous.

What I've tried is writing down all the definitions, but I can't see how they connect to provide a proof.

We know that $\forall \varepsilon>0$, $\forall x \in \mathbb{R}$, $\exists N$ such that $n \ge N$ implies $|f_n(x) - g(x)| < \varepsilon$. To show that $g$ is Lipschitz-continuous, I need to show that $\exists K>0$ such that $\forall x, z \in \mathbb{R}$, $|g(x) - g(z)| \le K|x-z|$. I can't seem to connect the pieces together and tie in the boundedness on the derivative to complete the proof.

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    $\begingroup$ Hint: are $f_n$ Lipschitz-continuous? $\endgroup$ – Adayah Aug 2 '17 at 9:18
  • $\begingroup$ The function is not pointwise convergent, the sequence $f_n$ is. $\endgroup$ – zhw. Aug 2 '17 at 17:31
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Step 1. Prove that each $f_n$ is Lipschitz continuous with the same constant $L=1$.

This is an easy consequence of the mean value theorem: given $x < y$, there exists $c\in (x,y)$ such that $$ |f_n(x) - f_n(y)| = |f_n'(c)| \cdot |x-y| \leq |x-y|. $$

Step 2. For every fixed $x,y\in\mathbb{R}$, by step 1 you have that $$ |f_n(x) - f_n(y)| \leq |x-y| \qquad \forall n\in\mathbb{N}. $$ Now it is enough to pass to the limit as $n\to +\infty$, obtaining $$ |g(x) - g(y)| \leq |x-y|. $$ (To be precise, you should use the triangular inequality: $$ \begin{split} |g(x) - g(y)| & \leq |g(x) - f_n(x)| + |g(y)-f_n(y)| + |f_n(x) - f_n(y)| \\ & \leq |g(x) - f_n(x)| + |g(y)-f_n(y)| + |x-y| \end{split} $$ and then pass to the limit as $n\to +\infty$.)

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  • $\begingroup$ Hmm, I'm still a bit confused, do you think you can provide a full proof based on your hints? Thank you very much. $\endgroup$ – TeTs Aug 2 '17 at 9:31
  • $\begingroup$ why the use of the triangle inequality? It is not enough to know that $|{\cdot}|$ is continuous in $\Bbb R$? $\endgroup$ – Masacroso Aug 2 '17 at 9:43
  • $\begingroup$ @Masacroso: yes, it is enough, but if you want to prove it you have to use the triangle inequality. $\endgroup$ – Rigel Aug 2 '17 at 9:44

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