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Let $\{u_\epsilon\}_{\epsilon>0}$ be a family of $C^1$ functions from $\mathbb{R}^n$ to $\mathbb{R}$ such that $\forall \epsilon$, $$|u_\epsilon| \le M \quad \text{ and } \quad |\nabla u_\epsilon| \le M$$ for a constant $M$. Then we can apply the Ascoli-Arzelà theorem and say that there exists a sequence $u_{\epsilon_i}$ that converges uniformly on compact subsets to a continuous limit function $u$.

Questions:

  1. Can we say that $u$ is Lipschitz continuous (or more)? Why?
  2. Can we say that $u_{\epsilon_i}$ converges to $u$ uniformly instead of just uniformly on compact subsets?
  3. Under which additional hypotehesis does the entire family $\{u_\epsilon\}$ converge to $u$?
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The set of functions with Lipschitz constant bounded by $M$ is $\bigcap_{x,y} \left\{f,|f(x)-f(y)|\le M||x-y|| \right\}$, i.e. an intersection of closed sets in compact-open topology. Thus, the limit of functions with gradient (and so Lipschitz constants) bounded by $M$ also has to be in this set. This should answer the first question in affirmative.

Consider $n=1$ and functions being the characteristic functions of $[-k,k]$, smoothened somehow (in the same way). Then this functions converge to the constant $1$ on the compacts, but not uniformly.

As for the third question, nothing valid comes to my head at the moment.

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