$f_{n}$ uniformly converges to $f$ and $|f'_{n} (x)| ≤ C $, is $f$ necessarily differentiable? Assume $f(x)$ is the uniform limit of real differentiable functions $f_{n}(x)$ on $[−1,1]$. Assume that $|f'_{n} (x)| ≤ C $ for some $C$ independent of $n$ and $x ∈ [−1, 1]$. Is the function $f(x)$ necessarily differentiable?
 A: No. Choose $f_n(x) := \sqrt{x^2 + \frac{1}{n}}$. Then, for all $x \in [-1, 1]$:
$$
\left \lvert \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \right \rvert = \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \leq \lvert x \rvert + \frac{1}{\sqrt{n}} - \lvert x \rvert = \frac{1}{\sqrt{n}} \overset{n \rightarrow \infty}{\longrightarrow}0
$$
Therefore
$$
\sup_{x \in [-1, 1]}\left \lvert \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \right \rvert \leq \frac{1}{\sqrt{n}} \overset{n \rightarrow \infty}{\longrightarrow}0
$$
and thus uniform convergence. The limit, i.e. the absolute value is not differentiable in $0$. Furthermore, for all $n \in \mathbb{N}$, $x \in [-1, 1]$:
$$
\lvert f_n'(x) \rvert = \left \lvert \frac{x}{\sqrt{x^2 + \frac{1}{n}}} \right \rvert = \frac{\sqrt{x^2}}{\sqrt{x^2 + \frac{1}{n}}} \leq \frac{\sqrt{x^2 + \frac{1}{n}}}{\sqrt{x^2 + \frac{1}{n}}} = 1
$$
A: For every $x,y \in [-1,1]$ and for every $n$ there exists $z_n \in (x,y)$ such that $f_n(x) - f_n(y) = f'(z_n)(x - y)$, so we have:
$$|f_n(x) - f_n(y)| \leq C |x - y|$$
for every $x,y \in [-1,1]$ and for every $n$, passing to the limit we get:
$$|f(x) - f(y)| \leq C |x - y|$$
for every $x,y \in [-1,1]$, so $f$ is Lipschitz and so is differentiable for $\mathcal{L}^1$ a.e. $x \in [-1,1]$.
But we can't conclude that $f$ is differentibale evreywhere, here there is a counterexample:
$$f_n(x) := (x^2)^{\frac{n}{2n -1}}$$
observe that $\lim_{n \to \infty}f_n(x) = |x|$ which is not differentiable at $x = 0$, the derivative is uniformly bounded wrt $n$ and the convergence is uniform
