Consider the sequence $\{g_n\}_{n \in \mathbb{N}}$ defined by
$$g_n(x) =\max \left\{ 0, \min \left\{1-\frac{n+1}{n}\Vert x\Vert, n- \ln\ln\left(1+\frac{1}{\Vert x\Vert}\right) \right\} \right\}.$$
How can I prove that $\{g_n\}_{n \in \mathbb{N}}$ converges to
$$g(x)= 1-\Vert x \Vert$$
in $W^{1,p}(B(0,r=1) \setminus 0)$, where $B(0,1) \subset \mathbb{R}^2$, for $$1 \le p \le 2$$ as $n \to \infty$?
That is, how can I prove that $$\int_{-1}^{1} |g_n(x)-g(x)|^p \ \mathrm{d}x \to 0 \quad \text{ and } \quad \int_{-1}^{1} |\nabla g_n(x)-\nabla g(x)|^p \ \mathrm{d}x \to 0$$ for $$1 \le p \le 2$$ as $n \to \infty$?
The one-dimensional case was solved by an answer given by @Dap. The following pictures refer to that case.
This is the plot of $g$ produced by Wolfram Alpha:
For $n=2$, this is the plot of $g_n$ produced by Wolfram Alpha:
My main difficulties are in estimating the integrals for $|x| \le |\tilde{x}|$, where $|\tilde{x}|$ is the value of $x$ where the maximum is achieved.
If a function $f$ is continuous, differentiable and monotone on an interval $[a,b]$ then $\int_a^b|\partial_xf|=|\int_a^b\partial_xf|=|f(b)-f(a)|.$
We can break up these functions into monotone pieces. $\int_{-1}^1|\partial_xg|=\int_{-1}^0|\partial_xg|+\int_{0}^1|\partial_xg|=2.$
Let $x^*$ be the value maximizing $g(x^*)$ subject to $x^*\in[0,1].$ Integrating $|\partial_xf|$ on the four intervals gives $\int_{-1}^1 |\partial_xg_n|=\int_{-1}^{-x^*} |\partial_xg_n|+\int_{-x^*}^0 |\partial_xg_n|+\int_{0}^{x^*} |\partial_xg_n|+\int_{x^*}^1 |\partial_xg_n|=4g_n(x^*)\approx 4.$
We have shown $\|\partial_x g_n\|_1\approx 4$ and $\|\partial_x g\|_1=2.$ So $\partial_x g_n$ does not converge to $\partial_x g$ in $L^1([-1,1]\setminus\{0\}).$ On a finite domain, $L^p$ convergence is a stronger requirement so they don't converge for $p>1$ either.
More generally, functions in $W^{1,1}$ have an (absolutely) continuous representative, and a sequence of functions $f_n$ with $f_n(0)=0$ cannot converge to a function with $f_n(0)=1$ in $W^{1,1}(0,1)$, or even in $BV$.