Let $U\subset\mathbb R^2$ be a bounded open set. Take $x^0\in \partial U$. Assume that there exist $r>0$ and $\gamma:\mathbb R\to\mathbb R$ of class $C^1$ such that $$U\cap B(x^0;r)=\{(x_1,x_2)\in B(x^0;r)\mid x_2>\gamma (x_1)\}.$$
Define $V=U\cap B(x^0;r/2)$.
Given $\lambda,\varepsilon>0$ and $x\in V$, define $x^\varepsilon=x^{\lambda,\varepsilon}=x+\lambda\varepsilon e_2$.
Acording to Evans book:
We can choose a sufficiently large number $\lambda>0$ such that $B(x^{\lambda,\varepsilon};\varepsilon)\subset U\cap B(x^0;r)$ for all $x\in V$ and all small $\varepsilon$.
I'm trying to get the bounds for $\varepsilon$ and $\lambda.$ According to this notes, the said inclusion holds for $\varepsilon <<1$ and $\lambda >>1$ but I can't see why.
Attempt. Let $x\in V$. Take $y\in B(x^{\lambda,\varepsilon};\varepsilon)$. Then $\|y-x^{\lambda,\varepsilon}\|<\varepsilon$ and thus $$\|y-x^0\|\leq \|y-x^{\lambda,\varepsilon}\|+\|(x-x^0)+\lambda\varepsilon e_2\|<\varepsilon+\frac{r}{2}+\lambda\varepsilon.$$ Therefore $y\in B(x^0;r)$ provided that $\varepsilon+\frac{r}{2}+\lambda\varepsilon<r$. However, it seems that there is a problem here because this estimate gives a lower bound for $\lambda$ instead of a upper bound. Namely, it gives $\lambda <\frac{r}{2(\varepsilon+1)}$ which is not compatible with $\lambda >>1$.
Any help would be appreciated.
There is a small mistake in your computation. $\varepsilon+\frac{r}{2}+\lambda\varepsilon<r$ provided $\varepsilon(1+\lambda)<\frac{r}{2}$. So you first choose $\lambda>1$ and then you can take $ \varepsilon<\frac{r}{2(1+\lambda)}$. To choose $\lambda$, take $y\in B(x^{\lambda,\varepsilon},\varepsilon)$. Then \begin{equation}y_2\ge x_2^\varepsilon-\varepsilon=x_2+\lambda\varepsilon-\varepsilon>\gamma(x_1)+\lambda\varepsilon-\varepsilon\ge \gamma(y_1)- \text{Lip}\gamma |x_1-y_1| +\lambda\varepsilon-\varepsilon \\ \ge \gamma(y_1)- \text{Lip}\gamma \varepsilon +\lambda\varepsilon-\varepsilon >\gamma(y_1) \end{equation} provided $\lambda>1+ \text{Lip}\gamma$.