= Solution
The <continuous dual space> of a real <normed vector space> $X$ is
$$
X^*=\{\phi:X\to\mathbb R:\phi\text{ is linear and bounded}\},
$$
with the <operator norm>
$$
\|\phi\|=\sup_{\|x\|\le1}|\phi(x)|.
$$
Two normed spaces are <isometric isomorphism of normed spaces>[isometrically isomorphic] when there is a bijective linear map between them that preserves norms.
For $y=(y_j)\in\ell^\infty$, define
$$
\Phi_y(x)=\sum_{j=1}^{\infty}x_jy_j,
\qquad x\in\ell^1.
$$
The series is absolutely convergent and
$$
|\Phi_y(x)|\le\|y\|_\infty\|x\|_1,
$$
so $\Phi_y\in(\ell^1)^*$ and $\|\Phi_y\|\le\|y\|_\infty$. For every $\varepsilon>0$, choose $j$ with $|y_j|>\|y\|_\infty-\varepsilon$ and test on $\operatorname{sgn}(y_j)e_j$. This gives the reverse inequality, hence
$$
\|\Phi_y\|=\|y\|_\infty.
$$
Conversely, given $\Phi\in(\ell^1)^*$, set $y_j=\Phi(e_j)$. Then $|y_j|\le\|\Phi\|$, so $y\in\ell^\infty$. For $x\in\ell^1$, its finite partial sums converge to $x$ in norm; continuity of $\Phi$ therefore gives
$$
\Phi(x)=\lim_{N\to\infty}\sum_{j=1}^Nx_j\Phi(e_j)
=\sum_{j=1}^{\infty}x_jy_j=\Phi_y(x).
$$
The representing sequence is unique, and the construction is linear and norm preserving. Thus the <duality of l1 and l infinity> proves
$$
\boxed{(\ell^1)^*\cong\ell^\infty\quad\text{isometrically}}.
$$
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