Let $\Omega=(-1,1)$ and $B=\{g \in W^{1,1}(\Omega) : \|g\|_{W^{1,1}(\Omega)}=\|g\|_{1,1}\leq 1\}$. I want to prove that $B$ is not compact with respect to the $L^1(\Omega)$ norm. As a hint, I am given the following sequence of functions: \begin{equation} f_n(x):=\begin{cases} 0 & x\leq 0 \\ nx & 0< x\leq 1/2n \\ 1/2 & x > 1/2n \end{cases} \end{equation}
My problem is that my limit function seems to be in $B$. Here's my solution:
a) First, I show that $(f_n) \subset B$
Let $n>0$. By definition, $\|f_n\|_{1,1}=\|f_n\|_{L^1}+\|f_n'\|_{L^1}$. WTS: $\|f_n\|_{1,1} \leq 1$.
We have that $\|f_n\|_{L^1}=\int_0^{1/2n}nx dx + \int_{1/2n}^1 \frac{dx}{2}=\frac{n}{2}\frac{1}{4n^2}+\frac{1}{2}(1-\frac{1}{2n})=\frac{4n-1}{8n}$.
Then, as $f_n'(x)=\begin{cases} n & \text{, if }x \in (0,1/2n)\\ 0 & \text{, if } x \in \Omega\setminus[0,1/2n] \\ \end{cases}$, I get that $\|f_n'\|_{L^1}=\int_0^{1/2n}ndx=\frac{1}{2}$ and hence $\|f_n\|_{1,1}=\frac{4n-1}{8n}+\frac{1}{2}=\frac{8n-1}{8n}\leq 1$.
b) Now, I show that the limit of $(f_n)$ is not in $W^{1,1}(\Omega)$ with respect to the topology induce by the $L^1(\Omega)$ norm, ie. $fn \xrightarrow[\text{}]{L^1} f$, but $f \not\in B$ :
Let $f(x):=\begin{cases} 0 & \text{, if }x \in (-1, 0]\\ 1/2 & \text{, otherwise.} \\ \end{cases}$. Then $(f-f_n)(x)=\begin{cases} 1/2 - nx & \text{, if }x \in (0,1/2n]\\ 0 & \text{, otherwise.} \\ \end{cases}$. Note that it is a positive function, for all $n>0$. Thus, $\|f-f_n\|_{L^1}=\int_0^{1/2n} \frac{1}{2}-nx dx= \frac{1}{2}\frac{1}{2n}-\frac{n}{2}\frac{1}{4n^2}=\frac{1}{4n}-\frac{1}{8n}=\frac{1}{8n}\rightarrow 0$, ie. $fn \xrightarrow[\text{}]{L^1} f$.
But since $f'=0$ almost everywhere: $\|f\|_{1,1}=\|f\|_{L^1}+\|f'\|_{L^1}=\|f\|_{L^1}=\int_0^1 \frac{dx}{2}=\frac{1}{2}\leq 1$, which is not what I am supposed to get since it shows that $f\in B$.
What did I do wrong? Thanks for the help.