Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 5 (Some Inequalities)

For conjugate exponents $p,q\in [1,\infty ]$, $1/p+1/q=1$, Hölder's inequality states that $\int |fg|\le \Vert f{\Vert }_p\Vert g{\Vert }_q$. If $r=\infty$ in the requested product estimate, every $p_i=\infty$ and multiplication of essential bounds proves it directly. Suppose $r<\infty$. Factors with $p_i=\infty$ can again be bounded pointwise by their essential suprema.

Here is an explicit finite-exponent proof of the remaining Generalized Holder inequality. Put $q_i=p_i/r$, so ${\sum }_{i}1/q_i=1$, and normalize $h_i=|f_i{|}^r/\Vert f_i{\Vert }_{p_i}^r$. If one norm is zero the product vanishes almost everywhere. Otherwise $\int h_i^{q_i}=1$. The weighted arithmetic-geometric mean inequality, applied to $h_i^{q_i}$ with weights $1/q_i$, givesIntegrating gives $\int {\prod }_i{h}_i\le 1$. If only one finite exponent remains, its weight is one and the same argument is an equality. Restoring the normalization and taking the $r$th root provesIn particular the product belongs to $L^r$.

For the projection estimate, use the three-dimensional functional Loomis-Whitney inequalityThis permitted three-dimensional result also follows directly by Cauchy-Schwarz inequality: integrate the square root of $g_1{g}_2$ in $x_3$, obtaining the square root of the product of their marginal integrals, and then apply Cauchy-Schwarz inequality in $(x_1,x_2)$ against $\sqrt{g_3}$.

Let $f_j$ be any nonnegative integrable function of the three coordinates excluding $x_j$. For $j=1,2,3$, define $G_j$ by integrating $f_j$ in $x_4$. First Hölder's inequality with three equal exponents yieldsThe fourth factor $f_4(x_1,x_2,x_3)$ is independent of $x_4$. A further Hölder's inequality with exponents $3$ and $3/2$, followed by the three-dimensional estimate, therefore givesThe marginal-integral identifications use Tonelli's theorem. Now take $f_j=1_{K_j}$. The projections are compact and hence measurable with finite measure. Every $x\in K$ has all its projected coordinates in the corresponding $K_j$, so $1_K(x)\le {\prod }_j{1}_{K_j}({\hat{x}}_j{)}^{1/3}$. ConsequentlyThis includes zero-measure projections and is sharp for axis-parallel rectangular boxes.

Read equality of the two positive series as equality of a finite total $M$, so the sequences are summable. This is the majorization of summable sequences setting. If equality is instead allowed to mean $\infty =\infty$, the assertion is false: the strictly decreasing sequences $x_i=1+1/(i+1)$ and $y_i=2+1/(i+1)$ satisfy every partial-sum inequality and have divergent equal extended totals. But a normalized nonnegative row of any doubly stochastic matrix averages numbers all at least $2$, and cannot give $x_i<2$. We prove the intended finite-total assertion constructively.

Since $y_j\to 0$ and $0<x_1\le y_1$, there is a finite $k$ such that $y_k\ge x_1>y_{k+1}$. PutThe two-coordinate stochastic averaging matrix sends $(y_k,y_{k+1})$ to $(x_1,u)$. Move the first of these values to the front by a finite permutation, leaving the other values in decreasing order. The tail iswhich is decreasing and strictly positive because $y_{k+1}\le u\le y_k$.

We need the residual majorization in order to repeat the construction. For $n<k$, the first $n$ values of $z$ are at least $x_1$, while every target-tail value is at most $x_1$. ThusFor $n\ge k$, the residual initial sum is the old initial sum with $x_1$ removed:The tail totals also agree, both equalling $M-x_1$. Therefore the same step applies to $(x_2,x_3,\dots )$ and $z$.

Iterating, let $P^{(m)}$ be the product of the first $m$ finite permutations and embedded averaging matrices. Each $P^{(m)}$ is a doubly stochastic matrix, each row is a finitely supported sequence, and the first $m$ entries of $P^{(m)}y$ are $x_1,\dots ,x_m$. Later operations act only on tail rows, so row $i$ becomes permanently fixed once $m\ge i$. Define $p_{ij}$ to be that stabilized row. ThenFor every fixed column $j$ and every $m$, its first $m$ stabilized entries sum to at most one, since they are entries of column $j$ in $P^{(m)}$. Consequently $s_j={\sum }_i{p}_{ij}\le 1$. By Tonelli's theorem,All $y_j$ are strictly positive and every summand in the last expression is nonnegative, so $s_j=1$ for every $j$. Thus both the row sums and the column sums are one, andThe finite-total argument is what prevents loss of column mass in the infinite construction. This completes the doubly stochastic realization of summable-sequence majorization.

A nonnegative sublinear operator satisfies $S(f+g)\le S(f)+S(g)$ and $S(cf)=|c|S(f)$ almost everywhere. It has weak type (1,1) if there is a finite constant $C$, independent of the function and threshold, such thatFor a signed or complex output use $|S(f)|$ in the level set. The bound controls the distribution of large values; it does not assert that $S(f)$ is integrable.

The domination hypothesis includes $r=0$. For any approximation error $h$, linearity and the triangle inequality therefore give the uniform estimateFor a general input $g$ and an approximating $f$, this impliesThe first term converges on the dense class, while the weak type (1,1) estimate controls the exceptional set where the second is large.

The printed symbol $E$ is not defined. Take $E=L^1(\mathbb{R})$, or more generally an $L^1$ subspace with its induced $L^1$ norm; density must be in that norm. Given $g\in E$, choose a sequence $f_n\in F$ withBy the assumed convergence on $F$, each $T_r{f}_n\to T_0{f}_n$ holds outside a null set $N_n$. Their countable union $N={\bigcup }_n{N}_n$ is still null. Thus outside $N$ the convergence statements for every member of this single approximation sequence hold simultaneously. No assertion of simultaneous convergence for the entire uncountable dense class is needed.

Use the approximants just chosen. The weak type (1,1) bound givesThe measures of these exceptional sets are summable. Indeed, the measure of their union for $n\ge m$ is at most $C{\sum }_{n\ge m}{2}^{-n}\to 0$. Hence their limsup is null; this is the Borel-Cantelli first lemma. Outside it, $S(g-f_n)\to 0$.

Remove also the countable null set from part (b). At each remaining point and for every $n$, the estimate in part (a) yieldsLetting $n\to \infty$ makes the right side zero. ThereforeThis proves almost-everywhere convergence from weak-type domination. The argument uses countably many measurable exceptional sets and does not assume that a supremum over all real parameters is measurable.

Let $U_{\alpha }=\{x:m_u(f)(x)>\alpha \}$. It is open: it is the union of all open intervals $I$ for which ${\int }_I|f|>\alpha |I|$. This also proves measurability of the uncentered maximal function. Triangle inequality inside each average proves sublinearity and absolute homogeneity.

Take a compact subset $K\subset U_{\alpha }$, and choose finitely many such intervals covering it. The one-dimensional Wiener covering lemma selects pairwise disjoint intervals $I_1,\dots ,I_N$ whose concentric triples cover $K$. For a finite family this follows by choosing a longest remaining interval and discarding all intervals intersecting it: each discarded interval has no greater length and lies in the triple of the chosen one. ThusTaking the supremum over compact subsets, by inner regularity of Lebesgue measure, givesThis establishes weak type (1,1). It is the one-dimensional uncentered maximal weak-type inequality for the finite measure $|f(t)|dt$; the constant need not be optimal for this question.

For $r>0$, define the forward and backward one-sided interval averaging operatorsand set $A_0^{\pm }f=f$. Each positive-radius average is bounded in absolute value by $m_u(f)(x)$. Although $x$ is an endpoint of its averaging interval, enlarge that interval slightly so that $x$ is interior, apply the maximal bound and let the enlargement shrink to zero.

To dominate the limiting identity too without assuming the differentiation result we are proving, use the sublinear operatorThe maximal estimate and Chebyshev inequality implyThus $S$ has weak type (1,1) and dominates every $A_r^{\pm }$, including $r=0$.

Smooth compactly supported functions are dense in $L^1$, obtained by truncating and then smoothing with a compactly supported approximate identity. For each such continuous function both averages tend to its value at every point. Applying part (ii) to the two families, and intersecting their full-measure convergence sets, gives $A_r^{\pm }f(x)\to f(x)$ almost everywhere.

The forward difference quotient of the primitive is $A_r^{+}f(x)$, and the backward quotient is $A_r^{-}f(x)$:They have the same limit, proving differentiability almost everywhere, with the stronger conclusionThis proves differentiation of an indefinite Lebesgue integral directly from the maximal estimate and density, without assuming $m_u(f)\ge |f|$ as an unproved prerequisite.

The interpolated exponent is strictly between $p_0$ and $p_1$. Apply Hölder's inequality to $|f{|}^p=|f{|}^{p(1-\theta )}|f{|}^{p\theta }$ with conjugate exponents $p_0/[p(1-\theta )]$ and $p_1/(p\theta )$. Their reciprocals sum to one, soThis gives $L^{p_0}\cap L^{p_1}\subseteq L^p$.

For the other inclusion take any $a>0$ and use the threshold decomposition between Lp spacesBecause $p_0-p<0<p_1-p$,Thus $g\in L^{p_0}$, $h\in L^{p_1}$ and $f=g+h$, proving $L^p\subseteq L^{p_0}+L^{p_1}$. Raising the two norm bounds to powers $1-\theta$ and $\theta$ gives a power of $a$ equal toand a power of $\Vert f{\Vert }_p$ equal to one. HenceFor later use, choose $a=\Vert f{\Vert }_p$ when $f\ne 0$. Each endpoint norm is then at most $\Vert f{\Vert }_p$, so $\Vert f{\Vert }_{L^{p_0}+L^{p_1}}\le 2\Vert f{\Vert }_p$ for the sum of Lp spaces norm. If $f=0$, take $g=h=0$.

Put $C=M_0^{1-\theta }{M}_1^{\theta }$. Since $p<\infty$, choose simple functions $f_n$ of finite-measure support with $f_n\to f$ in $L^p$. These belong to both endpoint domain spaces. The given simple-function estimate applies also to their differences, soBy completeness of Lp spaces, including $L^{\infty }$, there is $u\in L^q$ with $T{f}_n\to u$ in $L^q$. Taking limits of the simple-function bound yields $\Vert u{\Vert }_q\le C\Vert f{\Vert }_p$.

It remains to identify $u$ with the already defined $Tf$, rather than just define an unrelated extension. Write $X=L^{p_0}+L^{p_1}$ and $Y=L^{q_0}+L^{q_1}$ with their sum norms. The endpoint estimates implyIndeed, split $v=v_0+v_1$, bound the two images in their endpoint norms, and take the infimum over all splits. Part (i) gives $\Vert f_n-f{\Vert }_X\le 2\Vert f_n-f{\Vert }_p$, hence $T{f}_n\to Tf$ in $Y$.

The inclusion $L^q\subseteq Y$ is also continuous. For distinct finite endpoints this is the same threshold argument as part (i), after ordering the endpoints. If the larger endpoint is infinity, splitting at $a=\Vert v{\Vert }_q$ bounds both the lower endpoint norm of the large-value part and the supremum norm of the small-value part by $\Vert v{\Vert }_q$. If the endpoints coincide, the inclusion is immediate. Thus $T{f}_n\to u$ also in $Y$. Uniqueness of limits in the sum norm gives $u=Tf$ almost everywhere. Consequently the Riesz-Thorin theorem estimate extends as

Consider the complex-linear two-coordinate map $H(x,y)=(x+y,x-y)$. The triangle inequality shows that its operator norm from ${\ell }^1$ to ${\ell }^{\infty }$ is one; equality is attained at $(1,0)$. The complex parallelogram identityshows that its norm from ${\ell }^2$ to ${\ell }^2$ is $\sqrt{2}$. Interpolate these endpoint bounds by the Riesz-Thorin theorem with parameter $\eta =2/p$. ThenDividing the output norm by $2^{1/p}$ proves the scalar estimate:Apply it pointwise to $f,g$ and integrate its $p$th power. With $r=p/p^{\prime }=p-1>1$, this givesBy Minkowski inequality, the expression on the right is at mostTherefore the requested function inequality holds, namely the dual-exponent Clarkson inequality.

First use independent fair signs ${\epsilon }_i\in \{-1,1\}$, the Rademacher distribution convention for Bernoulli random variables in this setting. Let $\Omega =\{-1,1{\}}^d$ with uniform probability, write $s(\omega )={\sum }_i{\omega }_i{a}_i$ and $F(\omega )=\Vert s(\omega )\Vert$. We prove the sharp Rademacher second-moment inequality by a complete finite-cube calculation.

For a function on this cube define the coordinate-flip generator on a hypercubewhere ${\omega }^{(i)}$ flips just coordinate $i$. The Walsh characters ${\chi }_A(\omega )={\prod }_{i\in A}{\omega }_i$ form an orthonormal basis: distinct characters have expectation-zero product, and there are $2^d$ of them. Flipping a coordinate in $A$ changes its sign and other flips leave it unchanged, so $L{\chi }_A=-|A|{\chi }_A$. Expand the real function $F$ as ${\sum }_A\hat{F}(A){\chi }_A$. Since $F(-\omega )=F(\omega )$, changing variables $\omega \mapsto -\omega$ shows that $\hat{F}(A)=0$ whenever $|A|$ is odd. ThusThis derives the even-function spectral gap on a hypercube, rather than assuming it.

At each vertex with $s(\omega )\ne 0$, choose a norming functional ${\ell }_{\omega }$ of norm one satisfying ${\ell }_{\omega }(s(\omega ))=F(\omega )$, supplied by the Hahn-Banach theorem. For complex $E$ take real parts in the following inequalities. At a vertex with $s(\omega )=0$, use the zero functional. The supporting inequality for the norm givesBut $Ls=-s$, since every coefficient is a degree-one Walsh character. Summing the preceding inequalities with factor $1/2$ therefore proves $LF(\omega )\ge -F(\omega )$. As $F\ge 0$,Rearranging and taking square roots yieldsNo completeness of $E$ is needed. The constant is sharp: with two identical nonzero scalar coefficients, the absolute sum is zero with probability $1/2$ and twice the coefficient magnitude with probability $1/2$.

If Bernoulli instead means fair $\{0,1\}$ variables $b_i$, write $b_i=(1+{\epsilon }_i)/2$ and $a={\sum }_i{a}_i$. Add an independent fair sign ${\epsilon }_0$ and apply the proved estimate to $Y={\epsilon }_{0}a+{\sum }_i{\epsilon }_i{a}_i$. Conditional on either sign of ${\epsilon }_0$, $\Vert Y\Vert$ has the same distribution as $\Vert a+{\sum }_i{\epsilon }_i{a}_i\Vert$, by symmetry of the sign sum. Hence $\Vert Y{\Vert }_{L^p(E)}=2\Vert {\sum }_i{b}_i{a}_i{\Vert }_{L^p(E)}$ for $p=1,2$, proving the same bound in that convention.

The constant is documented in the primary sharp-constant study.

Fairness is essential. For a general $\{0,1\}$ Bernoulli random variable with success probability $0<t<1/2$, a single nonzero scalar coefficient gives $L^2/L^1=1/\sqrt{t}>\sqrt{2}$. Thus the printed term cannot be read as permitting arbitrary Bernoulli probabilities.