Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 5 / 2

One potential form of the Hardy–Littlewood–Sobolev inequality is: if $0<\alpha <n$, $1<p<n/\alpha$, and $1/r=1/p-\alpha /n$, thenA normalization constant for the Riesz potential only changes $C$. Equivalently, for $0<\lambda <n$, $p,s>1$ and $1/p+1/s+\lambda /n=2$,The exponent relation is necessary under dilation. We prove the potential estimate and then derive the bilinear version.

First establish the needed maximal estimate, so it is not an unproved step in the argument. Define the centered Hardy-Littlewood maximal function by $Mf(x)={\sup }_{t>0}|B_t{|}^{-1}{\int }_{B_t(x)}|f|$. For integrable $f$, cover a compact subset of $\{Mf>a\}$ by finitely many balls whose averages exceed $a$. Order these balls by decreasing radius and greedily retain a ball if it is disjoint from those already retained. Every discarded ball intersects a retained ball of at least its radius, so lies in its triple. HenceThe passage from compact subsets uses inner regularity of Lebesgue measure; the superlevel set is open because fixed-radius local averages are continuous. For $f\in L^p$, split $f=f{1}_{\{|f|>a/2\}}+f{1}_{\{|f|\le a/2\}}$. The second summand's maximal function is at most $a/2$, and the first is integrable. ThereforeThe layer cake representation and Tonelli theorem giveThus the Strong Lp bound for the Hardy-Littlewood maximal function is proved for every $p>1$.

For $I_{\alpha }|f|(x)$, split at radius $R$. Dyadic shells with radii $2^{-j}R$ giveFor the outer part, Hölder's inequality givesThe last integral is finite exactly because $p<n/\alpha$. Balance the two bounds by $R^{n/p}=\Vert f{\Vert }_p/Mf(x)$ to obtain the Hedberg inequalityThe zero cases follow directly or by a limit; $Mf$ is finite almost everywhere by the maximal estimate. Since $r(1-\theta )=p$, integration and the proved maximal bound yield $\Vert I_{\alpha }f{\Vert }_r\le C\Vert f{\Vert }_p$. Taking $\alpha =n-\lambda$ and pairing $I_{\alpha }|f|$ with $|g|\in L^{r^{\prime }}$ proves the bilinear form by Hölder's inequality, because $s=r^{\prime }$. All integrals are justified first for nonnegative truncations and then by monotone convergence and the resulting finite bound. The strong estimate does not include $p=1$ or $p=n/\alpha$.

Write $A_q=⟨f{⟩}_{w,q}$ and $N_q=\Vert f{\Vert }_{w,q}$. The exponent on the finite-set expression is $-1/q^{\prime }$ in the PDF; the TeX aid drops its prime. Ignore zero-measure sets, whose integrals vanish. The equivalence isFor the first bound, let $E_a=\{|f|>a\}$ and choose a finite-measure subset $E\subset E_a$. Since ${\int }_E|f|\ge a|E|$,Exhaust $E_a$ by its intersections with bounded balls. If its measure is finite these give the desired bound $a|E_a{|}^{1/q}$ in the limit; if it is infinite they force $N_q=\infty$. Now take the supremum over $a$.

For the other direction it suffices to assume $0<A_q<\infty$; the zero and infinite cases are immediate. For a set $E$ with $0<|E|=m<\infty$, the layer cake representation givesSplit at $t_0=A_q{m}^{-1/q}$. The first contribution is $m{t}_0$, and the second is $A_q^q{t}_0^{1-q}/(q-1)$. Their sum is $q^{\prime }{A}_q{m}^{1/q^{\prime }}$. Multiply by $m^{-1/q^{\prime }}$ and take the supremum. This proves the equivalence even when the quantities are extended-valued. The integral expression is an actual norm on the weak Lq space, since the triangle inequality holds inside every set integral. The distribution-function expression is generally a quasi-norm. The factor $q^{\prime }$ is sharp: $f(x)=x^{-1/q}$ on $(0,\infty )$ has $A_q=1$ and attains $N_q=q^{\prime }$ on every initial interval. In higher dimensions, multiply this example by the indicator of a unit box in the remaining coordinates.