Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 8 / 5

For $d\ge 2$, write ${\hat{x}}_j$ for the $(d-1)$ coordinates other than $x_j$, and setBy Tonelli theorem, $\Vert S_j{\Vert }_1=\Vert s_j{\Vert }_1$. The key estimate is the functional Loomis-Whitney inequality:Here is a proof in every dimension. In dimension two, the left-hand side separates into a product of two one-dimensional integrals, so Tonelli theorem gives equality. For $d\ge 3$, first integrate $x_d$. Apply Hölder's inequality with $d-1$ factors to $S_1,\dots ,S_{d-1}$, which may depend on $x_d$, and leave the factor $S_d$ outside that integral. WriteEach $H_j$ depends on the $d-2$ coordinates other than $j,d$. With $x^{\prime }=(x_1,\dots ,x_{d-1})$, the full integral is at mostPut $G(x^{\prime })={\prod }_{j<d}{H}_j^{1/(d-2)}$. A second application of Hölder's inequality, with exponents $d-1$ and $(d-1)/(d-2)$, bounds this expression byBy induction in dimension $d-1$, $\int G\le {\prod }_{j<d}(\int H_j{)}^{1/(d-2)}$. Since $\int H_j=\int S_j$, substitution proves the claimed functional inequality. Nonnegative integrands justify all integrations; truncation gives the extended-valued version if necessary.

The pointwise hypotheses imply $f^d\le {\prod }_j{S}_j$, soApply the proved functional inequality and take the $(d-1)/d$ power to obtainFor the useful case all the right-hand norms are finite. If one of them is zero, its line integral vanishes for almost every transverse coordinate and forces $f=0$ almost everywhere; there is no division by that factor. In dimension one the corresponding direct assertion is $\Vert f{\Vert }_{\infty }\le \Vert s_1{\Vert }_1$.

For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the fundamental theorem of calculus along the $j$th coordinate line givesTake $f=|u|$ and $s_j=|{\partial }_{j}u|$ in the introductory result. It follows thatThe last step is the arithmetic-geometric mean inequality. This bounds the target norm by the first-order Sobolev space norm. Approximate any $u\in W^{1,1}({\mathbb{R}}^d)=L_1^1({\mathbb{R}}^d)$ by compactly supported smooth functions, using cutoffs and mollification as in Question 4. The estimate applied to differences gives convergence in $L^{d/(d-1)}$, and subsequences of the simultaneous $L^1$ and target-norm convergence identify the two limits almost everywhere. Thus $W^{1,1}$ embeds continuously into $L^{d/(d-1)}$. The coordinate-derivative product bound also passes to the limit. For $d=1$, the line argument instead gives $\Vert u{\Vert }_{\infty }\le \Vert u^{\prime }{\Vert }_1$ and the embedding into $L^{\infty }$.

Each coordinate projection $S_j$ is compact, hence a Lebesgue measurable set, because it is the continuous image of the compact set $K$. Its $(d-1)$-dimensional measure is finite. Choose any nonnegative integrable function $\eta$ on $\mathbb{R}$ with integral one, and defineIf $x\in K$, every projection ${\hat{x}}_j$ belongs to $S_j$. If $x\notin K$, $f(x)=0$. Thus for every $j$, $f(x)\le 1_{S_j}({\hat{x}}_j)=\int s_j({\hat{x}}_j,t)dt$, so the introductory estimate applies. Tonelli theorem gives $\Vert s_j{\Vert }_1={\lambda }_{d-1}(S_j)$, andSubstitute these values and raise to the $d$th power:This proves the geometric Loomis--Whitney inequality, including zero-volume sets or zero-measure projections, without any division by their measures.