= Solution
Use the printed bilinear pairing $\phi_u(v)=\int uv$, which is complex linear in both arguments. For the <conjugate exponents> $p,q$, <Holder inequality> gives $|\phi_u(v)|\leq\|u\|_q\|v\|_p$, proving well-definedness, continuity, and $\|\phi_u\|\leq\|u\|_q$. Linearity of $u\mapsto\phi_u$ follows from linearity of integration.
For $u\ne0$, take
$$
v=\frac{\overline u\,|u|^{q-2}}{\|u\|_q^{q-1}},
$$
with the numerator defined as zero where $u=0$. Since $(q-1)p=q$, this has $\|v\|_p=1$ and $\phi_u(v)=\|u\|_q$. Hence $\boxed{\|\phi_u\|=\|u\|_q}$; 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 $w=f-h\ne0$. 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 functional
$$
\Psi(v)=\int |w|^{p-2}\overline w\,v
$$
vanishes on $K$. Since $\Lambda(w)=1$, each $v-\Lambda(v)w$ is in $K$, giving $\Psi(v)=\Lambda(v)\Psi(w)$ and $\Psi(w)=\|w\|_p^p$. Therefore
$$
\Lambda(v)=\int u\,v,\qquad
u=\frac{|w|^{p-2}\overline w}{\|w\|_p^p}\in L^q.
$$
Membership 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.
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