A special property of $\limsup$ in $\ell_1$ Let $w$ be any element in $\ell_1$, and $(w_n)$ be a bounded sequence in $\ell_1$ that converge to 0 pointwise. I want to prove 
 $$\limsup\limits_{n\to\infty}\|w_n-w\|_{1}=\limsup\limits_{n\to\infty}\|w_n\|_1+\|w\|_1$$
please help me to prove this . Thanks.
 A: This is basically continuation of the solution suggested in Daniel Fischer's comment.
I will use notation $(w^{(n)})$ for the sequence of elements of $\ell_1$. They are sequences, I will use $w^{(n)}_k$ for the $k$-th term of the $n$-th sequence.
We want to show $$\limsup_{n\to\infty} \|w^{(n)}-w\|_1 = \limsup_{n\to\infty} \|w^{(n)}\|_1 + \|w\|_1.$$

From triangle inequality we have
$$\newcommand{\norm}[1]{\|#1\|}
\norm{w^{(n)}-w} \le \norm{w^{(n)}}+\norm{w}.$$
This yields
$$\limsup_{n\to\infty} \norm{w^{(n)}-w} \le \limsup_{n\to\infty} (\norm{w^{(n)}}+\norm{w})
\le \\
\le \limsup_{n\to\infty} \norm{w^{(n)}}+ \limsup_{n\to\infty} \norm{w}
= \limsup_{n\to\infty} \norm{w^{(n)}}+ \norm{w}.$$
(Note that this part works for any norm, we did not use any special property of $\ell_1$-norm.)

Let $\newcommand{\ve}\varepsilon\ve>0$.
Since we know that $w\in\ell_1$, there exists $K$ such that
$$\newcommand{\abs}[1]{|#1|}
\sum_{k\ge K} \abs{w_k} < \frac\ve4.$$
Now the sequence $(w^{(n)})$ converges pointwise to $0$. This implies that for each $k\le K$ the sequence $(w^{(n)}_k)_{n=1}^\infty$ converges to $0$.1Therefore there exists $N$ such that
$$n\ge N, k\le K \qquad\Rightarrow\qquad \abs{w^{(n)}_k} < \frac\ve{4K}.$$
Now we can use these estimates and reverse triangle inequality to get
$$
\begin{align}
\norm{w^{(n)}-w}_1
&= \sum_{k=1}^\infty \abs{w^{(n)}_k - w_k}\\
&= \sum_{k<K} \abs{w^{(n)}_k - w_k} + \sum_{k\ge K} \abs{w^{(n)}_k - w_k} \\
&\ge \sum_{k<K} \abs{w_k} - \sum_{k<K} \abs{w^{(n)}_k} + \sum_{k\ge K} \abs{w^{(n)}_k} - \sum_{k\ge K} \abs{w_k} \\
&\ge \norm{w}_1 - \frac\ve4 -\frac\ve4+\norm{w^{(n)}}_1-\frac\ve4-\frac\ve4 \\
&= \norm{w}_1 + \norm{w^{(n)}}_1 - \ve
\end{align}
$$
for $n\ge N$. 
Now we can apply $\limsup$ on both sides to get
$$\limsup_{n\to\infty} \norm{w^{(n)}-w}_1 \ge \norm{w}_1 + \limsup_{n\to\infty} \norm{w^{(n)}}_1 - \ve.$$
Since this is true for every $\ve>0$, we get
$$\limsup_{n\to\infty} \norm{w^{(n)}-w}_1 \ge \norm{w}_1 + \limsup_{n\to\infty} \norm{w^{(n)}}_1.$$

1 In fact, it is true for each $k$. But since we want choose the same $N$ which works for several sequences, we can only take finitely many sequences.
