First, we define the notion of strong sub-differentiability(SSD) of a norm on a Banach space $X$. The norm $\Vert \cdot \Vert$ of $X$ is said to be SSD if the one-sided limit $$\lim_{t \to 0+} \frac{\Vert x+th\Vert - \Vert x \Vert}{t} $$ exists uniformly for $h \in S_X,$ and $x \in S_X=\{x \in X:\Vert x \Vert=1\}$. I have seen this result: "The norm of $\ell_1$ is SSD at every norm one norm attaining functional on $c_0$". A functional $f \in X^*$ attains its norm on $B_X$ if there exists $x_0 \in B_X$ such that $$ \langle f, x_0 \rangle =\sup \{ \langle f, y \rangle : y \in B_X \}. $$ Here, the set of all functionals in $X^∗$ that attain their norm is called the norm-attaining set, denoted by $NA((X, \Vert \cdot \Vert))$. Now, let us consider the mapping $T:\ell_1(\Gamma) \to \ell_2(\Gamma)$ and define a strictly convex dual norm $\Vert \cdot \Vert$ on $\ell_1(\Gamma)$ by $$ \Vert x \Vert=\Vert x \Vert_1+\Vert Tx \Vert_2,$$ for $x \in \ell_1(\Gamma)$. Here $\Gamma$ is an infinite index set. Now, if we consider $\ell_1(\mathbb{N})$, does this norm $\Vert \cdot \Vert$ is SSD at $f \in NA(c_0(\mathbb{N}))$. Kindly help me with my confusion. Thank you.
For reference, This new norm has been defined in [Book-Renormings in Banach spaces-pp-155 by Antonio J. Guirao, V. Zizler, and V. Montesinos].