= Solution
The <Hahn-Banach theorem> says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to\mathbb R$ is bounded and linear, then there is a bounded linear extension $G:X\to\mathbb R$ with $\|G\|=\|g\|$.
For $x\in X$, the map $\widehat x(f)=f(x)$ is linear on $X'$ and
$$
|\widehat x(f)|\leq\|f\|\,\|x\|,
$$
so $\widehat x\in X''$ and $\|\widehat x\|\leq\|x\|$. The assignment $Jx=\widehat x$ is visibly linear. If $x\ne0$, define $g$ on $\operatorname{span}\{x\}$ by
$$
g(\lambda x)=\lambda\|x\|.
$$
It has norm one, and Hahn--Banach extends it to an $f\in X'$ of norm one with $f(x)=\|x\|$. Therefore
$$
\|\widehat x\|\geq|\widehat x(f)|=\|x\|.
$$
Thus the <canonical embedding into the bidual> is an isometry and
$$
\boxed{\|x\|=\sup_{\|f\|\leq1}|f(x)|}.
$$
On $C([0,1])$, point evaluation $\ell(f)=f(0)$ has norm one for the $L^\infty$ norm, because a <continuous function>'s essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda\in(L^\infty)'$. If $\Lambda$ were represented by some $g\in L^1$, then
$$
f(0)=\int_0^1f(t)g(t)\,dt
$$
for every continuous $f$. Choose continuous $0\leq f_n\leq1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the <integral> makes the right side tend to zero, while the left side is always one, a contradiction. This <singular functional on L infinity> proves
$$
\boxed{(L^\infty)'\ne L^1(\mu).}
$$
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