Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 22H (Linear Analysis)

Let $q=p/(p-1)$. The scalar Young inequality $ab\le a^p/p+b^q/q$ follows by maximizing $ab-a^p/p$ over $a\ge 0$, obtaining $b^q/q$. If both norms are nonzero, apply it to $|x_j|/\Vert x{\Vert }_p$ and $|y_j|/\Vert y{\Vert }_q$, and sum over a finite initial segment. Passing to the increasing limit givesZero norms give the trivial case. This proves Hölder's inequality without assuming convergence of the product series in advance.

Apply Hölder's inequality to the finite sum with the second sequence $|x_j+y_j{|}^{p-1}$:Divide by the last power if the sum is nonzero. The resulting bound is independent of $N$, so passing to the limit proves both membership in ${\ell }^p$ and the Minkowski inequality

Assume $K$ is nonempty; without that necessary hypothesis the assertion is false. Put $d={\inf }_{z\in K}\Vert x-z{\Vert }_p$ and choose $y_n\in K$ with $\Vert x-y_n{\Vert }_p\to d$. Convexity gives $(y_n+y_m)/2\in K$, so its distance to $x$ is at least $d$. Apply the supplied uniform-convexity inequality to $x-y_n$ and $x-y_m$:Thus the minimizing sequence is Cauchy. Completeness of the Banach space ${\ell }^p$ gives $y_n\to y$; closedness gives $y\in K$, and continuity of the norm gives $\Vert x-y{\Vert }_p=d$. The minimum distance is attained. This is the nearest point in a uniformly convex Banach space property. The same inequality applied to two minimizers shows uniqueness, though existence alone was requested.