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Let's consider a family $\{u^\epsilon\}$ of $C^{1}$ functions on $\mathbb{R}^n$. Assume that we know (say, from Ascoli-Arzelà theorem) that there exists a subsequence that converges uniformly on compact sets to a limit function $u$. Then is $u$ $C^1$ or only Lipschitz (or less)? Why?

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The uniform limit of $C^1$ functions is ONLY continuous. Nothing more!

Let $u: \mathbb R\to\mathbb R$ a continous function, and let $\,j_\varepsilon(x)$ a mollifier, i.e., $\,j_\varepsilon\in C^\infty(\mathbb R)$,, $\,j_\varepsilon(x)\ge0$, $\,\int_{\mathbb R}j_\varepsilon=1\,$ and $\,\mathrm{supp}\,j_{\varepsilon}\subset [-\varepsilon,\varepsilon]$. Then $f_\varepsilon=f*j_\varepsilon\in C^\infty(\mathbb R)$ and $f_\varepsilon\to f$, uniformly on compact sets, as $\varepsilon\to 0$.

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Let $f:[0,1]\to \mathbb R$ be continuous . Extend $f$ to domain $\mathbb R$ by letting $f(x)=f(0)$ for $x<0$ and $f(x)=f(1)$ for $x>1.$

For $n\in \mathbb N,$ and integer $j$ with $0\leq j \leq n,$ let $g_n(j/n)=f(j/n),$ and let $g_n$ be continuous and linear on $[j/n,(j+1)/n]$ for $0\leq j\leq n-1.$ Let $g_n(x)=f(0)$ for $x<0$ and $g_n(x)=f(1)$ for $x>1.$

Now "round off the corners" of $g_n$ to obtain a continuously differentiable $h:\mathbb R\to \mathbb R$ with $|h_n(x)-g_n(x)|<1/n$ for all $x\in [0,1],$ and $h_n(x)=f(x)=f(1)$ for $x>1,$ and $h_n(x)=f(x)=f(0)$ for $x<0.$

Now $(g_n)_n$ converges uniformly to $f$ on $[0,1],$ so $(h_n)_n$ also does. And $h_n(x)=f(x)$ when $x\not \in [0,1].$ So $(h_n)_n$ converges uniformly to $f$ on $\mathbb R.$

Now $f$ restricted to $[0,1]$ may be any member of $C[0,1].$ In particular, $f$ restricted to $[0,1]$ may be nowhere differentiable and non-Lipschitz.

For example let $d:[0,1]\to [0,1]^2$ be the Peano space-filling curve and let $e((x,y))=x$ for $(x,y)\in [0,1]^2.$ Then $f=ed$ is continuous from $[0,1]$ to $[0,1],$ and for every $x\in [0,1]$ there is a sequence $(x_n)_{n\in \mathbb N}$ in $[0,1]$ \ $\{x\}$ with $\lim_{n\to \infty}|f(x_n)-f(x)|/|x_n-x|=\infty.$

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