Let $U \subset \mathbb{R}^n$ be open, bounded, and $C^1$. $U$ being $C^1$ means that, for any point $x^0 \in \partial U$, there exists $r > 0$ and a $C^1$ function $\gamma : \mathbb{R}^{n-1} \to \mathbb{R}$ such that:
$$ U \cap B(x^0. r) = \{x =(x_1, \dots, x_n) \in B(x^0, r) : x_n >\gamma(x_1, \dots, x_{n-1})\}. $$
Here, $B(y, R)$ represents the ball of radius $R > 0$ about the point $y \in \mathbb{R^n}$.
For the setting described above, define $V = U \cap B(x^0, \frac{r}{2})$. And for $x \in V, \lambda, \epsilon > 0$, define the shifted point $x^{(\epsilon, \lambda)} = x + \lambda \epsilon e_n$ ($e_i$ is the basis vector with $1$ in the $i$th coordinate and zeroes elsewhere).
I would like to show that there exist constants $M, N >0$ so that, if $x \in V$, $\lambda = M$, and $0 < \epsilon < N$, then $B(x^{(M, \epsilon)}, \epsilon) \subset U \cap B(x^0, r).$
If anyone is wondering where this question is coming from, this proposition is part of Theorem 3 on page 266 of Evans' PDE book, where he shows that $W^{k,p} (U)$ functions can be approximated in norm by functions in $C^\infty(\overline{U})$. I am trying to understand his proof and I am getting stuck on this claim.
From above, we know that $U \cap B(x^0. r) = \{x \in B(x^0, r) : x_n >\gamma(x_1, \dots, x_{n-1})\}$. And I'm really not sure how to pick $M, N$ so that a general point $z \in B(x^{(M, \epsilon)}, \epsilon)$ ($0 < \epsilon < N$) will satisfy the condition $z_n > \gamma(z_1, \dots, z_{n-1})$. It's easy to see that the center $x^{(M, \epsilon)}$ satisfies this condition since $(x^{(M, \epsilon)})_n = x_n + M\epsilon > x_n > \gamma(x_1 , \dots , x_{n-1}) =\gamma((x^{(M, \epsilon)})_1, \dots, (x^{(M, \epsilon)})_{n-1})$, but for the arbitrary point $z$ I am getting stuck.
Hints or solutions are greatly appreciated.
The key point is that $\gamma$ is a Lipschitz function. Indeed, only the values of $\gamma$ on some ball in $\mathbb R^{n-1}$ are relevant to the problem, and the norm of derivative of $\gamma$ has finite maximum there. Let $L$ be the Lipschitz constant of $\gamma$.
Let $x$ and $x^{(\epsilon,\lambda)}$ be as in your post. Consider any point $x'$ such that $|x'-x^{(\epsilon,\lambda)}|<\epsilon$. Then $$ x_n' > x_n + (\lambda-1)\epsilon>\gamma(x_1,\dots,x_{n-1})+ (\lambda-1)\epsilon $$ while $$ \gamma(x_1',\dots,x_{n-1}') \le \gamma(x_1,\dots,x_{n-1}) + L\epsilon $$ Therefore, choosing $\lambda=L+1$ works. As for $\epsilon$, it only needs to be small enough so that the shifted point does not out of $B(x^0,r)$. Having $\lambda \epsilon<r/2$ should prevent that.