Prove weak-weak-continouity. 
Let $p,q\in ]1,\infty[,$ $f:\mathbb{R}\to\mathbb{R}$ continuous, $|f(s)| \leq C|s|^{\frac{p}{q}}$, where $C>0$ is a constant.
Let $A:\ell^p \to \ell^q, (x_n)_{n\in\mathbb{N}}\mapsto (a_n+f(x_n))_{n\in\mathbb{N}},$ where $(a_n)_{n\in\mathbb{N}}\in\ell^q$.

I want to show that A is weak-weak-continous on the unit ball on $\ell^p$.

Here is what I have:
Let $(x_n^{(m)}) \rightharpoonup (x_n)$ on the unit ball, $(y_n) \in \ell^p$.
\begin{align*}
|\langle A((x_n^{(m)})) - A((x_n)), (y_n)\rangle|^q
& = |\langle (f(x_n^{(m)}) - f(x_n)), (y_n)\rangle|^q \\
& \le \langle (|f(x_n^{(m)})-f(x_n)|), (|y_n|)\rangle^q \\
& \le \langle (|f(x_n^{(m)})|+|f(x_n)|), (|y_n|)\rangle^q \\
& \le \sum_{n=1}^{\infty}|f(x_n^{(m)})y_n|^q + \sum_{n=1}^{\infty}|f(x_n)y_n|^q \\
& \le C\left(\sum_{n=1}^{\infty}|{x_n^{(m)}}^{\frac{p}{q}}y_n|^q + \sum_{n=1}^{\infty}|x_n^{\frac{p}{q}}y_n|\right)^q \\
& \le C\left(\sum_{n=1}^{\infty}|{x_n^{(m)}}^p y_n^q| + \sum_{n=1}^{\infty}|x_n^py_n^q|\right) \\
& \le C\left(\sum_{n=1}^{\infty}|y_n^q| + \sum_{n=1}^{\infty}|y_n^q|\right).
\end{align*}
But this doesn't give me what I want. I guess I'm totally wrong with this estimation.
Can you help me there?
 A: Suppose that $x^{(m)}\rightharpoonup x$ and $y\in\ell^{q'}$, where $q'$ is the conjugate exponent to $q$. Then, for any $N\in\mathbb N$, $$\begin{multline*}\left|\left<y,Ax^{(m)}-Ax\right>\right|=\left|\sum_{n=1}^{\infty}y_n(Ax_n^{(m)}-Ax_n)\right|=\left|\sum_{n=1}^{\infty}y_n(f(x_n^{(m)})-f(x_n))\right|\\
\leq\left|\sum_{n=1}^Ny_n(f(x_n^{(m)})-f(x_n))\right|+\left(\sum_{n=N+1}^{\infty}|y_n|^{q'}\right)^{1/q'}\left(\sum_{n=N+1}^{\infty}|f(x_n^{(m)})-f(x_n)|^q\right)^{1/q}.\end{multline*}$$ Note now that $(x^{(m)})_m$ is bounded in $\ell^p$, so $\|x^{(m)}\|_p\leq M$. Then, for any $N\in\mathbb N$, $$\begin{align*}\left(\sum_{n=N+1}^{\infty}|f(x_n^{(m)})-f(x_n)|^q\right)^{1/q}&\leq \left(\sum_{n=N+1}^{\infty}|f(x_n^{(m)})|^q\right)^{1/q}+\left(\sum_{n=N+1}^{\infty}|f(x_n)|^q\right)^{1/q}\\
&\leq \left(\sum_{n=N+1}^{\infty}\left|C|x_n^{(m)}|^{p/q}\right|^q\right)^{1/q}+\left(\sum_{n=N+1}^{\infty}\left|C|x_n|^{p/q}\right|^q\right)^{1/q}\\
&\leq C\|x^{(m)}\|_p^{p/q}+C\|x\|_p^{p/q}\leq CM^{p/q}+C\|x\|_p^{p/q},\end{align*}$$ therefore, for any $N\in\mathbb N$, $$\left|\left<y,Ax^{(m)}-Ax\right>\right|\leq \left|\sum_{n=1}^Ny_n(f(x_n^{(m)})-f(x_n))\right|+\left(\sum_{n=N+1}^{\infty}|y_n|^{q'}\right)^{1/q'}(CM^{p/q}+C\|x\|_p^{p/q}).$$ Let now $\varepsilon>0$, then there exists $N\in\mathbb N$ such that $$\left(\sum_{n=N+1}^{\infty}|y_n|^{q'}\right)^{1/q'}(CM^{p/q}+C\|x\|_p^{p/q})<\varepsilon,$$ therefore for this particular $N$, $$\left|\left<y,Ax^{(m)}-Ax\right>\right|\leq \left|\sum_{n=1}^Ny_n(f(x_n^{(m)})-f(x_n))\right|+\varepsilon.$$ Note now that, from weak convergence, $$x_n^{(m)}\xrightarrow[m\to\infty]{}x_n$$ for $n=1,\dots N$, therefore from continuity of $f$, $$f(x_n^{(m)})\xrightarrow[m\to\infty]{} f(x_n),$$ for $n=1,\dots N$. Therefore, there exists $m_0\in\mathbb N$ such that, for any $m\geq m_0$, $|f(x_n^{(m)})-f(x_n)|<N^{-1/q'}\varepsilon$, for $n=1,\dots N$, therefore $$\begin{align*}\left|\left<y,Ax^{(m)}-Ax\right>\right|&\leq\left|\sum_{n=1}^Ny_n(f(x_n^{(m)})-f(x_n))\right|+\varepsilon\leq \|y\|_{q'}\left(\sum_{n=1}^N|f(x_n^{(m)})-f(x_n)|^{q}\right)^{1/q}+\varepsilon\\
&\leq\|y\|_{q'}\varepsilon+\varepsilon.\end{align*}$$ This shows that $$\left<y,Ax^{(m)}-Ax\right>\xrightarrow[m\to\infty]{}0,$$ hence $Ax^{(m)}$ converges weakly to $Ax$.
