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Use the bilinear pairing $T_g(f)={\int }_{{\mathbb{R}}^n}fg$, so the map $g\mapsto T_g$ is complex linear. With the sesquilinear convention $\int f\overline{g}$ the representing map would instead be conjugate linear. For conjugate exponents $p,q$, Hölder's inequality gives $|T_g(f)|\le \Vert g{\Vert }_q\Vert f{\Vert }_p$. If $g\ne 0$, take $v=|g{|}^{q-2}\overline{g}$, defined as zero where $g=0$. ThenConsequently $\Vert T_g\Vert =\Vert g{\Vert }_q$. This also proves injectivity of the representation.

For surjectivity, let $\ell$ be a nonzero element of the continuous dual space of $L^p$ and choose $f_0$ with $\ell (f_0)=1$. Its kernel $K$ is a closed linear subspace. Use the allowed best-approximation fact to choose $h\in K$ and put $z=f_0-h$. Thus $\ell (z)=1$ andFor each $f$, the vector $f-\ell (f)z$ belongs to $K$. Therefore $\Phi (f)=\ell (f)\Vert z{\Vert }_p^p$, givingMembership follows from $(p-1)q=p$; the zero functional is represented by zero. This proves the isometric isomorphism $(L^p({\mathbb{R}}^n){)}^{\ast }\cong L^q({\mathbb{R}}^n)$ without using a representation theorem as a substitute for the proof. It is the closed-hyperplane proof of Lp duality.