Let $f\colon \ell_1\to \mathbf R$ be defined by $f((a_n)) = \sum_{n\geq 1} \frac{g(na_n)}{n}$, where $g\colon\mathbf R\to \mathbf R$ is defined by $g(x) = \mathrm{ln}(1+x^2)$.
I am trying to show that the map (derivative) $D\colon \ell_1\times \ell_1\to \mathbf R$ is $C^0$.If we define $Df((a_n),(b_n)) = \lim_{t\to 0}\frac{f\big((a_n)+t(b_n)\big) - f\big((a_n)\big)}{t}$is continuous. It is easy to see that $Df((a_n),(b_n))=\sum_{n\geq 1}b_ng'(a_n)$ just by using the limit. Moreover, by the mean value theorem, $|g'(x)|\leq 1$. Hence it follows easily that $|Df((a_n),(b_n))|\leq \|(b_n)\|_1$. Now $Df$ is linear in $(b_n)$ hence this boundedness implies continuity in the second argument. But I am stuck at making a concrete argument as to why it is jointly continuous.
EDIT: the question can be rephrased as follows:Let $g\colon \mathbf R\to \mathbf R$ is continuous (not necessarily linear) such that $|g(x)|\leq 1$ for all $x\in\mathbf R$. Is $F\colon \ell_1^2\to \mathbf R$, $F(a,b)= \sum_{n\geq 1} a_n g(b_n)$ continuous?
Any help is appreciated. Thanks!