$g\mapsto(f\mapsto\int fg)$ embeds $L^1$ isometrically in $(L^\infty)^*$; for Lebesgue measure on $[0,1]$ the embedding is not surjective.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space. For \(g\in L^1(\mu)\), the map \(\phi_g\colon L^\infty(\mu)\to\CC\), \(\phi_g(f)\coloneqq\int_X fg\dmu\), is a bounded linear functional with \(\norm{\phi_g}=\norm{g}_1\), so \(g\mapsto\phi_g\) is an isometric linear embedding \(L^1(\mu)\to(L^\infty(\mu))^*\) [@Fol13]. For Lebesgue measure on \([0,1]\) this embedding is not surjective.
The functional \(f\mapsto f(0)\) on the subspace \(C([0,1])\subseteq L^\infty([0,1])\) has norm \(1\), and by the Hahn–Banach theorem it extends to some \(\Lambda\in(L^\infty([0,1]))^*\). No \(g\in L^1([0,1])\) satisfies \(\Lambda=\phi_g\).