Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 9 / 1

The statement requires $K\ne \varnothing$; an empty set is closed and convex but contains no minimizer. Assume this necessary condition and put $d={\inf }_{g\in K}\Vert g-f{\Vert }_p$. Since $K$ is closed and $f\notin K$, $d>0$. Choose a minimizing sequence $g_j\in K$ with $\Vert g_j-f{\Vert }_p\to d$.

The midpoint $(g_j+g_k)/2$ lies in the convex set $K$. Applying the stated Clarkson inequality to $g_j-f$ and $g_k-f$ givesThe right side tends to zero, so $g_j$ is a Cauchy sequence. By completeness of Lp spaces, proved in Question 2(a), it converges to some $h$. Closedness puts $h$ in $K$, and continuity of the norm gives $\boxed{\Vert h-f{\Vert }_p=d}$. This is the closest point theorem for a closed convex subset of Lp. The same inequality applied to two minimizers forces their difference to be zero, so the minimizer is also unique.

For $g\in K$, convexity gives $h+t(g-h)\in K$ for $0\le t\le 1$. Thus $F(t)=\Vert h-f+t(g-h){\Vert }_p^p$ has a right minimum at $t=0$ and $F^{\prime }(0)\ge 0$. The derivative of a power of the Lp norm isConsequentlyThe printed hint has coefficient $2$ where the general coefficient is $p$. This does not affect the desired sign, since $p>0$. The complex conjugation is important for complex-valued functions; for real functions it can be omitted.

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.