 Solution

Use the printed bilinear pairing ${\varphi }_u(v)=\int uv$, which is complex linear in both arguments. For the conjugate exponents $p,q$, Holder inequality gives $|{\varphi }_u(v)|\le \Vert u{\Vert }_q\Vert v{\Vert }_p$, proving well-definedness, continuity, and $\Vert {\varphi }_u\Vert \le \Vert u{\Vert }_q$. Linearity of $u\mapsto {\varphi }_u$ follows from linearity of integration.

For $u\ne 0$, takewith the numerator defined as zero where $u=0$. Since $(q-1)p=q$, this has $\Vert v{\Vert }_p=1$ and ${\varphi }_u(v)=\Vert u{\Vert }_q$. Hence $\boxed{\Vert {\varphi }_u\Vert =\Vert u{\Vert }_q}$; the same identity holds for $u=0$. This also proves injectivity.

For surjectivity, let $\Lambda$ be a nonzero bounded linear functional on $L^p$. Choose $f$ with $\Lambda (f)=1$ and let $K=\ker \Lambda$. The closest point theorem for a closed convex subset of Lp supplies the nearest point $h\in K$; write $w=f-h\ne 0$. For every $z\in K$, the whole line $h+tz$ lies in $K$, so the norm derivative at the minimum vanishes for real $t$. Applying this also to $iz$ shows that the complex linear functionalvanishes on $K$. Since $\Lambda (w)=1$, each $v-\Lambda (v)w$ is in $K$, giving $\Psi (v)=\Lambda (v)\Psi (w)$ and $\Psi (w)=\Vert w{\Vert }_p^p$. ThereforeMembership follows from $(p-1)q=p$. The zero functional is represented by $u=0$. This proves the asserted Lp duality isometric isomorphism, including surjectivity, without a separate representation theorem.