Let $T$ be an element of $(\ell^\infty)^*$, then $T$ is linear and exists $C\ge 0$ such that $$\lvert Tu\rvert\le C\lVert u\rVert _\infty,\quad \forall u\in\ell^\infty.$$ In particular we have $$\lvert Tu\rvert \le C\lVert u\rVert _\infty,\quad\forall u\in c_0,$$ that is the restriction $T_0:=T_{|c_0}$ is an element of $(c_0)^*$ and $\lVert T_0\rVert_{c_0^*}\le C.$ We know that $c_0^*\simeq \ell^1$, namely exists $y\in \ell^1$ such that $$T_u=T_0u=\sum_{k=1}^\infty u_ky_k,\quad\forall u\in c_0,$$ where $y_k=T{e_k}$ for all $k\in \mathbb{N}$ being $e_k$ the kth element of the basis.
The operator $T_0\colon c_0\to\mathbb{R}$ has two extensions: the first is $T$ and the second is the operator $\overline{T}\colon\ell^\infty\to \mathbb{R}$ defined by $$\overline{T}u=\sum_{k=1}^\infty u_ky_k,\quad\forall u\in\ell^\infty.$$
Not sure what $T=\overline{T}$.
For example let $S$ the bounded linear operator, defined on $c$, the space of all convergent real sequence, by $$Su=\lim_{k\to\infty} u_k,\quad\forall u\in c.$$
For Hahn-Banach Theorem, let $T_S$ the bounded linear extension of $S$ to $\ell^\infty$: for this operator the restriction $T_0$ is the zero operator on $c_0$: $$T_0u=Su=\lim_{k\to\infty} u_k=0\quad\forall u\in c_0.$$
So the operator $T_0=0\colon c_0\to\mathbb{R}$ has two distinct extensions to $\ell^\infty$: the first is the operator $\overline{T}=0$ and the second is $T_S$.
From this fact I would like the following thing:
There is no isomorphism between $(\ell^\infty)^*$ and $\ell^1\simeq c_0^*$, because such operator $j\colon (\ell^\infty)^*\to c_0^*$ it could not be injective: would have $j(\overline{T})=j(0)=0$ and $j(T_S)=0$.
Question. Why $$j(T_S)=0?$$
One could ask the question in two versions, which are not the same:
are $(\ell^\infty)^*$ and $\ell^1$ isomorphic?
are $(\ell^\infty)^*$ and $\ell^1$ canonically isomorphic, in the sense that the embedding of $\ell^1$ into its double dual is surjective?
In general, it is possible that the answer to the second question is no, even though the answer to the first question is yes.
That's not the case here, though. You have that $\ell^1$ is separable while $(\ell^\infty)^*$ is not (since $\ell^\infty$ is not separable). So there cannot be any isomorphism as Banach spaces. In particular, your $j$ cannot be injective.