The functional ${\mathrm{\Lambda }}_g$ has norm $\mathrm{\parallel }g{\mathrm{\parallel }}_q$; for $q=\mathrm{\infty }$ assume $\mu$ is semifinite

The upper bound $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\le \mathrm{\parallel }g{\mathrm{\parallel }}_q$ is exactly [L1].

If $g=0$ in $L^q(\mu )$, then ${\mathrm{\Lambda }}_g=0$ by definition, so $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=0=\mathrm{\parallel }g{\mathrm{\parallel }}_q$. Hence only the case $g\mathrm{\ne }0$ remains.

Assume $1<p<\mathrm{\infty }$. Choose a representative $u$ of $g$ and define $s(x):=\{\begin{array}{ll}\overline{u(x)}\mathrm{/}\mathrm{\mid }u(x)\mathrm{\mid },& u(x)\mathrm{\ne }0,\\ 0,& u(x)=0,\end{array}f(x):=\frac{\mathrm{\mid }u(x){\mathrm{\mid }}^{q-1}s(x)}{\mathrm{\parallel }g{\mathrm{\parallel }}_q^{q-1}}\mathrm{.}$ Then $\mathrm{\mid }f{\mathrm{\mid }}^p=\mathrm{\mid }u{\mathrm{\mid }}^q\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_q^q$ by [L2], so $\mathrm{\parallel }[f]{\mathrm{\parallel }}_p^p=\frac{1}{\mathrm{\parallel }g{\mathrm{\parallel }}_q^q}\int \mathrm{\mid }u{\mathrm{\mid }}^{q}d\mu =1.$ Also $fu=\mathrm{\mid }u{\mathrm{\mid }}^q\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_q^{q-1}$, so ${\mathrm{\Lambda }}_g([f])=\frac{1}{\mathrm{\parallel }g{\mathrm{\parallel }}_q^{q-1}}\int \mathrm{\mid }u{\mathrm{\mid }}^{q}d\mu =\mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$ Therefore $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\ge \mathrm{\parallel }g{\mathrm{\parallel }}_q$. Together with step 1.1, this gives $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=\mathrm{\parallel }g{\mathrm{\parallel }}_q$.

Assume $p=1$, so $q=\mathrm{\infty }$, and assume $\mu$ is semifinite. Put $M:=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Step 2.1 leaves only $g\mathrm{\ne }0$, so $M>0$. Choose a representative $u$ of $g$, and define $s(x):=\{\begin{array}{ll}\overline{u(x)}\mathrm{/}\mathrm{\mid }u(x)\mathrm{\mid },& u(x)\mathrm{\ne }0,\\ 0,& u(x)=0.\end{array}$ For $0<\epsilon <M$, [L3] gives a measurable set $E_{\epsilon }:=\{\mathrm{\mid }u\mathrm{\mid }>M-\epsilon \}$ of positive measure. By [L4], choose $F_{\epsilon }\subseteq E_{\epsilon }$ with $0<\mu (F_{\epsilon })<\mathrm{\infty }$, and set $f_{\epsilon }:=\frac{s{\mathbf{1}}_{F_{\epsilon }}}{\mu (F_{\epsilon })}\mathrm{.}$ Then $\mathrm{\parallel }[f_{\epsilon }]{\mathrm{\parallel }}_1=1$ and ${\mathrm{\Lambda }}_g([f_{\epsilon }])=\frac{1}{\mu (F_{\epsilon })}{\int }_{F_{\epsilon }}\mathrm{\mid }u\mathrm{\mid }d\mu \ge M-\epsilon \mathrm{.}$ Hence $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\ge M-\epsilon$ for every $0<\epsilon <M$, so $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }\ge M=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Step 1.1 gives the reverse inequality, and therefore $\mathrm{\parallel }{\mathrm{\Lambda }}_g\mathrm{\parallel }=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Step 2.2 proves the strict-exponent case, and step 3.1 proves the $q=\mathrm{\infty }$ endpoint under semifiniteness. Together with steps 1.1 and 2.1, this proves the proposition.