If $(X,\mathfrak{M},\mu)$ is a $\sigma$ -finite measure-space and $p,q$ are Hölder conjugates with $p<\infty$ , then we show that $L^q$ is isometrically isomorphic to the dual space of $L^p$ .

For any $g\in L^q$ , define the linear map \begin{equation*} \Phi_g\colon L^p\rightarrow\mathbb{C},\ f\mapsto \Phi_g(f)=\int fg\,d\mu. \end{equation*}This is a bounded linear map with operator norm $\Vert\Phi_g\Vert=\Vert g\Vert_q$ (see $L^p$ -norm is dual to $L^q$ ), so the map $g\mapsto\Phi_g$ gives an isometric embedding from $L^q$ to the dual space of $L^p$ . It only remains to show that it is onto.

So, suppose that $\Phi\colon L^p\rightarrow\mathbb{C}$ is a bounded linear map. It needs to be shown that there is a $g\in L^q$ with $\Phi=\Phi_g$ . As any $\sigma$ -finite measure is equivalent to a probability measure, there is a bounded $h>0$ such that $\int h\,d\mu=1$ . Let $\tilde\Phi\colon L^\infty\rightarrow\mathbb{C}$ be the bounded linear map given by $\tilde\Phi(f)=\Phi(hf)$ . Then, there is a $g_0\in L^1$ such that \begin{equation*} \Phi(hf)=\tilde\Phi(f)=\int fg_0\,d\mu \end{equation*}for every $f\in L^\infty$ (see bounded linear functionals on $L^\infty$ ). Set $g=h^{-1}g_0$ and, for any $f\in L^p$ , let $f_n$ be the sequence \begin{equation*} f_n = f 1_{\{|h^{-1}f|<n\}}. \end{equation*}As $h^{-1}f_n\in L^\infty$ , \begin{equation*} \Vert f_ng\Vert_1 =\Vert h^{-1}f_ng_0\Vert_1=\Phi(\operatorname{sign}(fg_0)f_n)\le\Vert\Phi\Vert \Vert f_n\Vert_p. \end{equation*}Letting $n$ tend to infinity, dominated convergence says that $f_n\rightarrow f$ in the $L^p$ -norm, so Fatou's lemma gives \begin{equation*} \Vert f g\Vert_1\le\liminf_{n\rightarrow\infty}\Vert f_n g\Vert_1\le\Vert\Phi\Vert\Vert f\Vert_p. \end{equation*}In particular, $\Vert g\Vert_q\le\Vert\Phi\Vert$ (see $L^p$ -norm is dual to $L^q$ ), so $g\in L^q$ . As $|f_ng|\le|fg|$ are in $L^1$ , dominated convergence finally gives \begin{equation*} \int fg\,d\mu =\lim_{n\rightarrow\infty}\int f_ng\,d\mu=\lim_{n\rightarrow\infty}\Phi(f_n)=\Phi(f) \end{equation*}so $\Phi_g=\Phi$ as required.