 Solution

TakeIt is evaluation against the first coordinate vector of $c_0$, hence continuous for $\sigma ({\ell }^1,c_0)$. The preceding continuous-dual result says that a functional continuous for $\sigma ({\ell }^1,c)$ must have coefficient sequence $({\gamma }_{\infty },{\gamma }_1,{\gamma }_2,\dots )$ for a convergent sequence $\gamma$. To represent ${\alpha }_1$, it would require ${\gamma }_{\infty }=1$ but every ${\gamma }_j=0$, impossible.

Directly, $e_1-e_n\to 0$ in $\sigma ({\ell }^1,c)$ because its pairing is ${\gamma }_{\infty }-{\gamma }_{n-1}\to 0$, whereas ${\psi }_1(e_1-e_n)=1$ for $n\ge 2$.