2.9 Fourier analysis and incidence of complex lines99FBE9B23B3C

Proof. Let $f\in \mathcal S$. We first prove the second rule in (2.9.3). By the mean value theorem, the difference quotients of $\xi \mapsto \mathrm e^{-2\pi \mathrm i x\cdot \xi }$ in the direction of $\xi _ j$ are bounded by $2\pi |x_ j|$, and $x_ jf$ is integrable. Hence dominated convergence allows us to differentiate under the integral sign: $\partial _{\xi _ j}\widehat f$ exists and is the transform of $-2\pi \mathrm i x_ jf$, which is continuous. Since $x_ jf\in \mathcal S$, induction shows that $\widehat f$ is smooth and that $\partial ^\beta \widehat f$ is the transform of $(-2\pi \mathrm i x)^\beta f$.

For the first rule, Fubini's theorem allows us to integrate in $x_ j$ before the other coordinates. For almost every value of the other coordinates, $f$ and $\partial _ jf$ are integrable on the corresponding line, and an integration by parts over $[-a,a]$ followed by $a\to \infty $ gives

because the boundary terms are bounded by $|f(x)|$ at $x_ j=\pm a$, and $f(x)\to 0$ as $|x|\to \infty $. Integrating over the other coordinates proves the first rule. Combining the two rules, we find that $\xi ^\alpha \partial ^\beta \widehat f$ is a constant multiple of the transform of $\partial ^\alpha (x^\beta f)$. This function belongs to $\mathcal S$ and is therefore integrable, so $\xi ^\alpha \partial ^\beta \widehat f$ is bounded by a multiple of its $L^1$ norm. Hence $\widehat f\in \mathcal S$.

For the Gaussian, consider first $d=1$, put $g(x)=\mathrm e^{-\pi x^2}$, which belongs to $\mathcal S(\mathbb R)$, and set $F=\widehat g$. Since $-2\pi \mathrm i xg=\mathrm i g'$, the two rules give $F'=\mathrm i\, \widehat{g'}=\mathrm i(2\pi \mathrm i\xi )F=-2\pi \xi F$, so $F(\xi )\mathrm e^{\pi \xi ^2}$ is constant. Moreover $F(0)=1$: squaring the positive integral $F(0)$ and using polar coordinates in the plane, where $s_1=2\pi $, gives $F(0)^2=2\pi \int _0^\infty r\mathrm e^{-\pi r^2}\, \mathrm d r=1$. Hence $F(\xi )=\mathrm e^{-\pi \xi ^2}$. In dimension $d$, Fubini's theorem writes the transform of $\mathrm e^{-\pi |x|^2}$ as a product of one-dimensional transforms, which proves the formula for $t=1$. For general $t$, the linear substitution $x=y/\sqrt t$ gives

Proof. Let $f\in \mathcal S$ and $\varepsilon >0$. Insert the factor $\mathrm e^{-\pi \varepsilon ^2|\xi |^2}$ into the inversion integral. By Fubini's theorem and lemma 2.9.2 with $t=\varepsilon ^2$,

The functions $g_\varepsilon $ are positive with integral one, and the substitution $y=\varepsilon y'$ shows that their mass outside a fixed ball $B(0,\delta )$ tends to zero as $\varepsilon \to 0$. Since $f$ is bounded, and uniformly continuous because its gradient is bounded, splitting the convolution integral into the regions $|y|<\delta $ and $|y|\geq \delta $ shows that $f*g_\varepsilon \to f$ uniformly. On the left-hand side, $\widehat f\in \mathcal S\subset L^1$ by lemma 2.9.2, so dominated convergence lets $\varepsilon \to 0$ under the frequency integral. This proves the inversion formula.

Write $\mathcal F$ for the Fourier transform on $\mathcal S$, $\mathcal F^*$ for the inverse integral transform $\mathcal F^*g(x)=\int g(\xi )\mathrm e^{2\pi \mathrm i x\cdot \xi }\, \mathrm d\xi $, and $Rf(x)=f(-x)$. The substitution $x\mapsto -x$ shows that $\mathcal F^*=R\mathcal F=\mathcal FR$ on $\mathcal S$. In particular $\mathcal F^*$ maps $\mathcal S$ into itself, and both compositions $\mathcal F^*\mathcal F$ and $\mathcal F\mathcal F^*$ equal $R\mathcal F\mathcal F$, which is the identity on $\mathcal S$ by the inversion formula. For $f,g\in \mathcal S$, applying inversion to $g$ and Fubini's theorem, we obtain Parseval's identity

the interchange of integrals is justified because the double integral of the absolute value is at most $\| f\| _1\| \widehat g\| _1$.

By lemma 2.5.1(5), applied to Lebesgue measure, $C_ c^\infty (\mathbb R^ d)$, and hence $\mathcal S$, is dense in $L^2$. By Parseval's identity, $\mathcal F$ maps sequences in $\mathcal S$ that are Cauchy in $L^2$ to Cauchy sequences, and $L^2$ is complete; therefore $\mathcal F$ extends uniquely from $\mathcal S$ to a bounded operator on $L^2$, and this extension is an isometry. Since $\mathcal F^*=R\mathcal F$, the inverse integral transform extends in the same way. Both compositions of the two extensions are continuous and equal to the identity on the dense subspace $\mathcal S$, hence everywhere. Thus the extension is unitary, with the stated inverse.

Finally let $f\in L^1\cap L^2$. We choose $f_ j\in C_ c^\infty $ converging to $f$ both in $L^1$ and in $L^2$: first replace $f$ by $f\mathbf1_{B(0,R)}$ with $R$ large, and then mollify; by lemma 2.5.1(2) the mollifications converge in both norms, since their supports lie in a fixed compact set. The integral transforms of the $f_ j$ converge uniformly to that of $f$, because the difference is bounded by $\| f_ j-f\| _1$, while by isometry they converge in $L^2$ to the $L^2$ transform of $f$. A subsequence of an $L^2$ convergent sequence converges almost everywhere: choose it so that the squared $L^2$ errors are summable, and apply Tonelli's theorem to the sum of the squared errors. Hence the two transforms of $f$ agree almost everywhere. $\square$

Proof. Let $g\in L^2$ be the inverse $L^2$ transform of $\widehat\mu $, so that $\widehat g=\widehat\mu $. For $\phi \in \mathcal S$ we have

The first equality follows from the inversion formula for $\phi $ and Fubini's theorem. For the second, we apply Parseval's identity to $\phi $ and $\overline g$; it extends from $\mathcal S$ to $L^2$ because the $L^2$ transform is unitary. We also use that the transform of $\overline g$ at $\xi $ is $\overline{\widehat g(-\xi )}$; this holds for Schwartz functions by definition, and it extends to $L^2$ because the maps $g\mapsto \widehat{\overline g}$ and $g\mapsto \overline{\widehat g(-\, \cdot \, )}$ are continuous from $L^2$ to $L^2$.

Consequently, for every test function $\phi $ the number $\int \phi g\, \mathrm d x=\int \phi \, \mathrm d\mu $ is real when $\phi $ is real and nonnegative when $\phi \geq 0$. By the Cauchy–Schwarz inequality on compact sets, $g$ is locally integrable, so it defines a positive distribution. By proposition 2.5.3(3) its mollifications $g*\rho _\varepsilon $ are nonnegative, and they converge to $g$ in $L^1$ on compact sets by lemma 2.5.1(2). Hence $g$ is real and nonnegative almost everywhere. The measures $\mu $ and $g\, \mathrm d x$ are Radon by proposition 2.2.2, and they define the same distribution; by proposition 2.5.3(4), $\mu =g\, \mathrm d x$. In particular $\int g\, \mathrm d x=\mu (\mathbb R^ d)<\infty $, and $\| g\| _2=\| \widehat\mu \| _2$ because the $L^2$ transform is unitary. $\square$

Proof. For $z\in \mathbb R^ d$, the substitution $v=\pi t|z|^2$ gives

and both sides are $+\infty $ when $z=0$. By lemma 2.9.2 with $t$ replaced by $1/t$, we have $\mathrm e^{-\pi t|z|^2}=t^{-d/2}\int \mathrm e^{-\pi |\xi |^2/t}\mathrm e^{-2\pi \mathrm i z\cdot \xi }\, \mathrm d\xi $. Taking $z=x-y$ and integrating with respect to $\mu \times \mu $, we obtain

Fubini's theorem applies here because the Gaussian in $\xi $ is integrable and $\mu \times \mu $ is finite. Now insert the integral formula for $|x-y|^{-s}$ into $I_ s(\mu )$. All integrands are nonnegative, so Tonelli's theorem allows us to integrate first in $(x,y)$, then use the last identity, and finally integrate in $t$ before $\xi $, even when the integrals are infinite:

For $\xi \neq 0$, the substitution $v=\pi |\xi |^2/t$ evaluates the inner integral as

and the point $\xi =0$ is a Lebesgue null set. Multiplying the constants gives the formula. $\square$

Proof. By lemma 2.9.2 with $\xi =0$ and $t=1$, the Gaussian $\mathrm e^{-\pi |x|^2}$ has integral one. Integrating in polar coordinates and substituting $v=\pi r^2$, we obtain

which proves (2.9.8). By lemma 2.9.2, $\widehat\mu (\xi )=\mathrm e^{-\pi |\xi |^2}$. Hence theorem 2.9.6 and polar coordinates give

The substitution $v=2\pi r^2$ evaluates the radial integral as $\Gamma (s/2)/(2(2\pi )^{s/2})$. Inserting the values of $b_{d,s}$ and $s_{d-1}$ proves the formula. $\square$

Proof. The multiplier is real and bounded by one in absolute value, so Plancherel's theorem shows that $T_{ab}$ is self-adjoint with norm at most one. For $v\in C_ c^\infty $, the first rule in (2.9.3) (lemma 2.9.2), applied twice, gives $\widehat{-\Delta v}(\xi )=4\pi ^2|\xi |^2\widehat v(\xi )$ and $\widehat{\partial _ a\partial _ bv}(\xi )=-4\pi ^2\xi _ a\xi _ b\widehat v(\xi )$. Multiplying the first identity by the multiplier of $T_{ab}$ gives the second, which proves (3).

To construct the kernel we use the heat kernel $H_ t(z)=(4\pi t)^{-d/2}\mathrm e^{-|z|^2/(4t)}$, $t>0$. By lemma 2.9.2, $\widehat H_ t(\xi )=\mathrm e^{-4\pi ^2t|\xi |^2}$, and hence $\widehat{\partial _ a\partial _ bH_ t}(\xi )=-4\pi ^2\xi _ a\xi _ b\mathrm e^{-4\pi ^2t|\xi |^2}$. For $0<\varepsilon <R<\infty $ put

Since $\partial _ a\partial _ bH_ t(z)=\bigl (z_ az_ b/(4t^2)-\delta _{ab}/(2t)\bigr )H_ t(z)$, where $\delta _{ab}$ is the Kronecker symbol, we have

Hence $\sup _ z|\partial _ a\partial _ bH_ t(z)|\leq C_ dt^{-d/2-1}$, and the substitution $z=\sqrt t\, w$ gives $\int |\partial _ a\partial _ bH_ t(z)|\, \mathrm d z\leq C_ dt^{-1}$. Therefore the kernel $k_{\varepsilon ,R}$ is bounded and integrable, and Fubini's theorem, applied with the integrable majorant $|\partial _ a\partial _ bH_ t(z)|$ on $\mathbb R^ d\times [\varepsilon ,R]$, gives its Fourier transform:

Let $f\in L^1\cap L^2$. Then $k_{\varepsilon ,R}*f$ is bounded and integrable, hence in $L^2$, and by Fubini's theorem and the last assertion of theorem 2.9.4 its Fourier transform is $\widehat{k_{\varepsilon ,R}}\widehat f$. As $\varepsilon \downarrow 0$ and $R\uparrow \infty $, these multipliers are bounded by one and converge to $-\xi _ a\xi _ b/|\xi |^2$ for $\xi \neq 0$. By dominated convergence on the Fourier side and Plancherel's theorem, $k_{\varepsilon ,R}*f\to T_{ab}f$ in $L^2$.

For $z\neq 0$ define

Since $H_ t(z)=t^{-d/2}H_1(z/\sqrt t)$, every derivative satisfies

and induction on $|\gamma |$ shows that $\partial ^\gamma H_1(w)=P_\gamma (w)\mathrm e^{-|w|^2/4}$ for a polynomial $P_\gamma $, of degree $m_\gamma $ say. Fix $0<c\leq c'$, and let $c\leq |z|\leq c'$ and $|\gamma |\geq 2$. Since $|P_\gamma (w)|\leq C_\gamma (1+|w|)^{m_\gamma }$ and $\mathrm e^{-|z|^2/(4t)}\leq \mathrm e^{-c^2/(4t)}$, we obtain $|\partial ^\gamma H_ t(z)|\leq g_\gamma (t)$, where

This function is integrable on $(0,\infty )$ and does not depend on $z$: near $t=0$ the factor $\mathrm e^{-c^2/(4t)}$ dominates every power of $t^{-1}$, and for $t\geq 1$ the exponent satisfies $(d+|\gamma |)/2\geq 3/2>1$. By the mean value theorem and dominated convergence, we may therefore differentiate under the integral sign any number of times. Hence $K_{ab}$ is smooth on $\mathbb R^ d\setminus \{ 0\} $, and its derivatives are the integrals of the corresponding derivatives of $\partial _ a\partial _ bH_ t$. Moreover, the integrals $\int _0^\varepsilon $ and $\int _ R^\infty $ of $|\partial _ a\partial _ bH_ t(z)|$ tend to zero as $\varepsilon \downarrow 0$ and $R\uparrow \infty $, uniformly for $c\leq |z|\leq c'$.

Differentiating the identity $H_{r^2s}(rz)=r^{-d}H_ s(z)$ twice in $z$ and substituting $t=r^2s$, we find that $K_{ab}(rz)=r^{-d}K_{ab}(z)$ for $r>0$. Thus $K_{ab}$ is homogeneous of degree $-d$ and $\nabla K_{ab}$ is homogeneous of degree $-d-1$, and the bounds in (1) follow from the boundedness of the continuous functions $K_{ab}$ and $\nabla K_{ab}$ on the unit sphere. The same scaling shows that the function $z\mapsto \int _0^\infty |\partial _ a\partial _ bH_ t(z)|\, \mathrm d t$ is homogeneous of degree $-d$; it is bounded on the unit sphere by $\int _0^\infty g_\gamma (t)\, \mathrm d t$, where $\partial ^\gamma =\partial _ a\partial _ b$ and $c=c'=1$, so it is at most $C_ d|z|^{-d}$.

To prove (2), let $f$ and $F$ be as there. By the Cauchy–Schwarz inequality on $F$, the function $f$ is integrable, so $f\in L^1\cap L^2$. Let $L$ be a compact set disjoint from $F$. For $x\in L$,

where the outer integral may be taken over $F$. The differences $x-y$, with $x\in L$ and $y\in F$, form a compact set not containing the origin, on which $c\leq |x-y|\leq c'$ for some $0<c\leq c'$. By the uniform estimates just proved, the inner integral converges to $K_{ab}(x-y)$ uniformly on this set, and it is bounded by $\int _0^\infty |\partial _ a\partial _ bH_ t(x-y)|\, \mathrm d t\leq C_ dc^{-d}$. Hence, since $f$ is integrable, $k_{\varepsilon ,R}*f$ converges uniformly on $L$ to $\int K_{ab}(x-y)f(y)\, \mathrm d y$. Since it also converges to $T_{ab}f$ in $L^2$, a subsequence converges almost everywhere to $T_{ab}f$, and the two limits agree almost everywhere on $L$. Covering the open set $\mathbb R^ d\setminus F$ by countably many such compact sets proves (2). For $d=1$ the kernel vanishes, in accordance with $T_{11}=-1$: a direct differentiation gives $\partial _ z^2H_ t=\bigl (z^2/(4t^2)-1/(2t)\bigr )H_ t=\partial _ tH_ t$, so $K_{11}(z)=\lim _{t\uparrow \infty }H_ t(z)-\lim _{t\downarrow 0}H_ t(z)=0$ for $z\neq 0$. $\square$

Proof. We use the Calderón–Zygmund decomposition of $f$ at height $\lambda $ into a bounded part $g$, handled by the $L^2$ bound for $T_{ab}$, and parts $b_ Q$ of mean zero on disjoint cubes $Q$, handled by lemma 2.9.10. The dyadic cubes are the cubes $2^ k(m+[0,1)^ d)$ with $k\in \mathbb Z$ and $m\in \mathbb Z^ d$. Two dyadic cubes are either disjoint or nested, and each dyadic cube $Q$ lies in a unique dyadic cube of twice its side length, its parent, which has volume $2^ d|Q|$. Call a dyadic cube $Q$ selected if the average of $|f|$ over $Q$ exceeds $\lambda $, while the average over every strictly larger dyadic cube containing $Q$ does not. Every dyadic cube on which the average of $|f|$ exceeds $\lambda $ is contained in a selected cube, because the averages over its successive ancestors are at most $\| f\| _1$ divided by their volumes and therefore tend to zero. The selected cubes are pairwise disjoint. For each of them, the average over its parent is at most $\lambda $, and hence

Moreover, $|f|\leq \lambda $ almost everywhere outside the union of the selected cubes. Indeed, let $x$ be a point outside this union at which the conclusion of theorem 2.8.3 holds. Every dyadic cube containing $x$ has average of $|f|$ at most $\lambda $, since otherwise it would lie in a selected cube. The dyadic cube of side $2^ k$ containing $x$ lies in the cube $x+(-2^ k,2^ k)^ d$, whose volume is $2^ d$ times larger. Hence the average of $|f-f(x)|$ over it tends to zero as $k\to -\infty $, so its average of $|f|$ tends to $|f(x)|$, and therefore $|f(x)|\leq \lambda $.

For a selected cube $Q$, let $f_ Q=|Q|^{-1}\int _ Qf$ be the average of $f$. Define $g=f$ outside the selected cubes and $g=f_ Q$ on each selected cube $Q$, and put $b_ Q=(f-f_ Q)\mathbf1_ Q$. Each $b_ Q$ has integral zero, and

By the Cauchy–Schwarz inequality, $|Q||f_ Q|^2\leq \int _ Q|f|^2$, so $\| b_ Q\| _2\leq 2\| f\mathbf1_ Q\| _2$. Since the $b_ Q$ have disjoint supports, the series $\sum _ Qb_ Q$ converges to $f-g$ in $L^2$.

Let $c_ Q$ be the center of $Q$, and let $Q^*$ be the cube with center $c_ Q$ and $4\sqrt d$ times the side length of $Q$. If $x\notin Q^*$ and $y\in Q$, then $|x-c_ Q|\geq 2\operatorname {diam}Q$ and $|y-c_ Q|\leq \operatorname {diam}Q/2$, so every point of the segment from $x-c_ Q$ to $x-y$ has norm at least $|x-c_ Q|/2$. The function $b_ Q\in L^2$ vanishes outside the compact set $\overline Q$, which does not meet $\mathbb R^ d\setminus Q^*$, and it has integral zero. Therefore parts (1) and (2) of lemma 2.9.10 and the mean value theorem give, for almost every $x\notin Q^*$,

Integrating in polar coordinates about $c_ Q$ over the region $|x-c_ Q|\geq 2\operatorname {diam}Q$, which contains $\mathbb R^ d\setminus Q^*$, we obtain

Put $B^*=\bigcup _ QQ^*$. Then $|B^*|\leq (4\sqrt d)^ d\sum _ Q|Q|\leq C_ d\lambda ^{-1}\| f\| _1$. Enumerate the selected cubes, and let $S_ N$ be the partial sums of $\sum _ Qb_ Q$. Since $S_ N\to f-g$ in $L^2$, also $T_{ab}S_ N\to T_{ab}(f-g)$ in $L^2$, and a subsequence converges almost everywhere. Moreover $\int _{\mathbb R^ d\setminus B^*}|T_{ab}S_ N|\leq \sum _ Q\int _{\mathbb R^ d\setminus Q^*}|T_{ab}b_ Q|\leq C_ d\| f\| _1$, so Fatou's lemma gives

Finally, $T_{ab}f=T_{ab}g+T_{ab}(f-g)$, and the set where $|T_{ab}f|>\lambda $ is contained in the union of $B^*$, the set where $|T_{ab}g|>\lambda /2$, and the set of $x\notin B^*$ with $|T_{ab}(f-g)(x)|>\lambda /2$. By Chebyshev's inequality and the $L^2$ bound for $T_{ab}$, the second set has measure at most $4\lambda ^{-2}\| g\| _2^2\leq 2^{d+2}\lambda ^{-1}\| f\| _1$, and the third has measure at most $2\lambda ^{-1}\int _{\mathbb R^ d\setminus B^*}|T_{ab}(f-g)|\leq C_ d\lambda ^{-1}\| f\| _1$. Together with the bound for $|B^*|$, this proves the lemma. $\square$

Proof. Assume first that $1<p<2$, and let $f\in L^ p\cap L^2$. Fix $\lambda >0$ and split $f=f_1+f_2$ with $f_1=f\mathbf1_{\{ |f|>\lambda \} }$ and $f_2=f\mathbf1_{\{ |f|\leq \lambda \} }$. Since $|f_1|\leq \lambda ^{1-p}|f|^ p$, we have $f_1\in L^1\cap L^2$, and $f_2\in L^2$. Apply lemma 2.9.11 to $f_1$, and Chebyshev's inequality with the $L^2$ bound to $f_2$, both with threshold $\lambda /2$. This gives

For a measurable function $h$, Tonelli's theorem applied to $|h(x)|^ p=\int _0^{|h(x)|}p\lambda ^{p-1}\, \mathrm d\lambda $ expresses $\| h\| _ p^ p$ as the integral of $p\lambda ^{p-1}$ times the distribution function $\lambda \mapsto |\{ |h|>\lambda \} |$, also when the norm is infinite. We apply this to $h=T_{ab}f$: multiply the last inequality by $p\lambda ^{p-1}$ and integrate in $\lambda $. By Tonelli's theorem again, the first term on the right contributes $C_ d\int |f|\int _0^{|f|}p\lambda ^{p-2}\, \mathrm d\lambda \, \mathrm d x=C_ dp\| f\| _ p^ p/(p-1)$, and the second contributes $C_ d\int |f|^2\int _{|f|}^\infty p\lambda ^{p-3}\, \mathrm d\lambda \, \mathrm d x=C_ dp\| f\| _ p^ p/(2-p)$; the inner integrals converge precisely because $p>1$ and $p<2$. Therefore

For $p=2$ the bound is Plancherel's theorem.

Now let $2<p<\infty $, put $p'=p/(p-1)\in (1,2)$, let $f\in L^ p\cap L^2$, and write $h=T_{ab}f\in L^2$. For integers $N\geq 1$, put $E_ N=B(0,N)\cap \{ |h|\leq N\} $ and $g_ N=\mathbf1_{E_ N}h|h|^{p-2}$. Then $|g_ N|\leq N^{p-1}\mathbf1_{E_ N}$, so $g_ N\in L^2\cap L^{p'}$, and $|g_ N|^{p'}=\mathbf1_{E_ N}|h|^ p$. Since $T_{ab}$ is self-adjoint on $L^2$, the case $p'$ already proved and Hölder's inequality give

The integral on the left is finite. If it is nonzero, dividing by its $(1/p')$th power gives $(\int _{E_ N}|h|^ p)^{1/p}\leq C_{d,p'}\| f\| _ p$; if it is zero, this bound is trivial. The sets $E_ N$ increase to $\{ |h|<\infty \} $, which has full measure since $h\in L^2$, so monotone convergence gives $\| T_{ab}f\| _ p\leq C_{d,p'}\| f\| _ p$.

By lemma 2.5.1(5), $C_ c^\infty \subset L^ p\cap L^2$ is dense in $L^ p$, and $L^ p$ is complete, so $T_{ab}$ has a unique bounded extension from $L^ p\cap L^2$ to $L^ p$. Finally, for $v\in C_ c^\infty $, lemma 2.9.10(3) gives $\| \partial _ a\partial _ bv\| _ p=\| T_{ab}\Delta v\| _ p\leq C_{d,p}\| \Delta v\| _ p$, and summing over $a$ and $b$ proves (2.9.13). $\square$

Proof. (1) For $k=0$ the first assertion holds because $\chi $ is bounded on compact sets. Let $k\geq 1$ and $1\leq j\leq d$. For a test function $\phi $ on $U$, the function $\chi \phi $ is a test function with $\partial _ j(\chi \phi )=\chi \partial _ j\phi +\phi \partial _ j\chi $, so

Thus $\partial _ j(\chi v)=\chi \partial _ jv+(\partial _ j\chi )v$. The functions $\partial _ jv$ and $v$ belong to $W^{k-1,p}_{\mathrm{loc}}(U)$, so by induction on $k$ both terms belong to $W^{k-1,p}_{\mathrm{loc}}(U)$; hence $\chi v\in W^{k,p}_{\mathrm{loc}}(U)$, and iterating the formula for first derivatives gives the Leibniz formula, exactly as for smooth functions.

Now let $\operatorname{Supp}\chi $ be a compact subset of $U$, and let $W$ be as in the statement. By the Leibniz formula, $\chi v$ and its derivatives of order at most $k$ vanish almost everywhere outside $\operatorname{Supp}\chi $, and there they are bounded by $C\sum _{|\beta |\leq k}|\partial ^\beta v|$, where $C$ depends on $d$, $k$ and the derivatives of $\chi $ of order at most $k$. Hence their $L^ p$ norms are at most $C\| v\| _{W^{k,p}(W)}$, which is finite when $W\Subset U$. To identify the derivatives of the extension by zero, let $\phi $ be a test function on $\mathbb R^ d$, and choose a smooth cutoff function $\zeta $ for $\operatorname{Supp}\chi $ in $U$ which equals $1$ on a neighborhood $N$ of $\operatorname{Supp}\chi $ (lemma 2.5.1(4)). Then $\zeta \phi $ is a test function on $U$, and $\partial ^\alpha (\zeta \phi )=\partial ^\alpha \phi $ on $N$. Since $\chi v$ and $\partial ^\alpha (\chi v)$ vanish almost everywhere outside $\operatorname{Supp}\chi $,

This proves (1).

(2) The composite $v\circ \Phi $ is defined almost everywhere: if $Z\subset U$ is a Borel null set, the change of variables formula for the diffeomorphism $\Phi ^{-1}$ gives $|\Phi ^{-1}(Z)|=\int _ Z|\det D\Phi ^{-1}(x)|\, \mathrm d x=0$. For a smooth function $w$ on $U$ and $1\leq |\alpha |\leq k$, the classical chain rule gives

with smooth functions $c_{\alpha \beta }$ on $V$ that depend only on $\Phi $: for $|\alpha |=1$ this is $\partial _ j(w\circ \Phi )=\sum _ i\partial _ j\Phi _ i\, (\partial _ iw)\circ \Phi $, and differentiating the formula once more, with the product rule and the chain rule applied to each $(\partial ^\beta w)\circ \Phi $, gives the induction step.

Assume first that $v$ vanishes almost everywhere outside the compact set $K$. By (1), applied with a smooth cutoff function equal to $1$ near $K$, the extension of $v$ by zero belongs to $W^{k,p}(\mathbb R^ d)$. Its derivatives of order at most $k$ vanish almost everywhere outside $K$, because $v$ vanishes almost everywhere on the open set $\mathbb R^ d\setminus K$ and hence so do its distributional derivatives. Choose $\delta >0$ such that the compact set $K'=\{ x:\operatorname {dist}(x,K)\leq \delta \} $ lies in $U$. For $\varepsilon <\delta $, the mollification $v_\varepsilon =v*\rho _\varepsilon $ is smooth and vanishes outside $K'$, and $\partial ^\beta v_\varepsilon =(\partial ^\beta v)*\rho _\varepsilon \to \partial ^\beta v$ in $L^ p(\mathbb R^ d)$ for $|\beta |\leq k$, by the remark preceding the lemma, applied on $\mathbb R^ d$ to a bounded open set containing $K'$, outside of which both sides vanish.

Let $J$ be the maximum of $|\det D\Phi ^{-1}|$ on $K'$, and let $M\geq 1$ bound the functions $|c_{\alpha \beta }|$ on the compact set $\Phi ^{-1}(K')$. By the change of variables formula,

for every measurable function $g$ on $U$ that vanishes outside $K'$. The smooth functions $w_\varepsilon =v_\varepsilon \circ \Phi $ vanish outside $\Phi ^{-1}(K')$. Applying this inequality to $v_\varepsilon -v$ and its derivatives, we find that $w_\varepsilon \to v\circ \Phi $ and, for $1\leq |\alpha |\leq k$,

with $\| G_\alpha \| _ p\leq MJ^{1/p}\sum _{1\leq |\beta |\leq |\alpha |}\| \partial ^\beta v\| _ p$. For a test function $\phi $ on $V$, we obtain

Hence $\partial ^\alpha (v\circ \Phi )=G_\alpha \in L^ p(V)$. These functions, and $v\circ \Phi $, vanish almost everywhere outside $\Phi ^{-1}(K)$, and summing the bounds over $|\alpha |\leq k$ gives $\| v\circ \Phi \| _{W^{k,p}(V)}\leq C\| v\| _{W^{k,p}(U)}$, with $C$ depending only on $k$, $p$, $J$ and $M$, that is, on $k$, $p$, $\Phi $ and $K$ once $\delta $ has been fixed in terms of $K$ and $U$.

In general, let $V'\Subset V$ be open. Choose a smooth cutoff function $\chi $ for the compact set $\Phi (\overline{V'})$ in $U$ (lemma 2.5.1(4)). By (1), $\chi v\in W^{k,p}(U)$, and it vanishes outside $\operatorname{Supp}\chi $; by the case just treated, $(\chi v)\circ \Phi \in W^{k,p}(V)$. Since $(\chi v)\circ \Phi =v\circ \Phi $ on $V'$, we obtain $v\circ \Phi \in W^{k,p}(V')$. Hence $v\circ \Phi \in W^{k,p}_{\mathrm{loc}}(V)$. $\square$

Proof. Let $e_ a$ be the $a$th coordinate vector and $h>0$. Taylor's formula with integral remainder gives

The $L^ p$ norm is invariant under translation. Moreover, $\omega (t)\, \mathrm d t=2h^{-2}(h-t)\, \mathrm d t$ is a probability measure on $[0,h]$, so Hölder's inequality gives $|\int _0^ hg\, \omega \, \mathrm d t|^ p\leq \int _0^ h|g|^ p\omega \, \mathrm d t$ for every measurable function $g$ on $[0,h]$. Applying this to $g(t)=\partial _ a^2v(x+te_ a)$, integrating in $x$ and using Tonelli's theorem, we obtain

Hence $h\| \partial _ av\| _ p\leq 2\| v\| _ p+(h^2/2)\| \partial _ a^2v\| _ p$. Since $|\nabla v|\leq \sum _ a|\partial _ av|$ and $|\partial _ a^2v|\leq |D^2v|$, summing over $a$ yields $\| \nabla v\| _ p\leq 2dh^{-1}\| v\| _ p+(dh/2)\| D^2v\| _ p$. The choice $h=2\varepsilon /d$ proves the lemma with $C_{d,\varepsilon }=d^2/\varepsilon $. $\square$

Proof. We prove the estimate for smooth $v$, pass to general $v$ by mollification, and then apply the result to first derivatives. Assume first that $v$ is smooth on $B$. Let $Q\subset B$ be a ball of radius $r$, let $v_ Q=|Q|^{-1}\int _ Qv$ be the average of $v$ over $Q$, and let $x\in Q$. We claim that

For $z\in Q$, write $z=x+s\theta $ with $s=|z-x|$ and $\theta \in S^{d-1}$; then $|v(z)-v(x)|\leq \int _0^ s|\nabla v(x+t\theta )|\, \mathrm d t$, and $|v(x)-v_ Q|$ is at most the average over $z\in Q$ of these segment integrals. Since $Q$ is convex and open and contains $x$, for each $\theta $ the radii $s>0$ with $x+s\theta \in Q$ form an interval $0<s<R(\theta )$, and $R(\theta )\leq 2r$. In polar coordinates about $x$ ((3.1.2) on $B(x,2r)\supset Q$), the average of the segment integrals is

Interchanging the integrals in $s$ and $t$ produces the factor $|Q|^{-1}\int _ t^{R(\theta )}s^{d-1}\, \mathrm d s$, which is at most $(2r)^ d/(d|Q|)$ and hence bounded by a constant depending only on $d$. The remaining integral $\int _{S^{d-1}}\int _0^{R(\theta )}|\nabla v(x+t\theta )|\, \mathrm d t\, \mathrm d S(\theta )$ is the integral on the right side of the claim, written in polar coordinates. This proves the claim.

By Hölder's inequality with $p'=p/(p-1)$, the right side of the claim is at most $C_{d,p}r^{1-d/p}\| \nabla v\| _{L^ p(Q)}$, because

and this integral is finite exactly when $(1-d)p'>-d$, that is, when $p>d$. (For $d=1$ the exponent $(1-d)p'$ is zero.)

Let $\delta >0$ be the distance from $B'$ to $\mathbb R^ d\setminus B$. If $x,y\in B'$ and $0<|x-y|<\delta /2$, the ball $Q=B(x,2|x-y|)$ lies in $B$ and contains $x$ and $y$. Applying the preceding bound at both points gives

For $x\in B'$, the same bound on the ball $Q=B(x,\delta /2)\subset B$, together with the estimate $|v_ Q|\leq |Q|^{-1/p}\| v\| _{L^ p(B)}$ from Hölder's inequality, gives $\sup _{B'}|v|\leq C\| v\| _{W^{1,p}(B)}$ with $C$ depending on $d,p,\delta $. If $|x-y|\geq \delta /2$, then $|v(x)-v(y)|/|x-y|^\alpha \leq 2(\delta /2)^{-\alpha }\sup _{B'}|v|$. These bounds prove the estimate for smooth $v$ on $B'$, and hence on $\overline{B'}$, since $v$ is continuous on $B\supset \overline{B'}$.

For general $v\in W^{1,p}(B)$, choose a ball $B''$, depending only on $B'$ and $B$, with $B'\Subset B''\Subset B$, and let $v_\varepsilon =v*\rho _\varepsilon $, which is smooth on $B''$ for small $\varepsilon $. By the remark preceding lemma 2.9.14, $\partial _ av_\varepsilon =(\partial _ av)_\varepsilon $, and $v_\varepsilon \to v$ and $\partial _ av_\varepsilon \to \partial _ av$ in $L^ p(B'')$. The estimate for smooth functions, applied on $B'\Subset B''$ to the differences of the $v_\varepsilon $, shows that they form a Cauchy family in the norm of $C^{0,\alpha }(\overline{B'})$. Their uniform limit is continuous on $\overline{B'}$, agrees almost everywhere with $v$, and satisfies the asserted estimate, because $\| v\| _{W^{1,p}(B'')}\leq \| v\| _{W^{1,p}(B)}$. Two continuous functions that agree almost everywhere on an open set agree everywhere on it. Hence the continuous representatives obtained from different balls $B'\Subset B$ agree on their overlaps and define a single continuous representative on $B$.

Finally, let $v\in W^{2,p}_{\mathrm{loc}}(U)$. On every ball $B\Subset U$, both $v$ and its first derivatives $\partial _ av$ belong to $W^{1,p}(B)$. By the preceding paragraph, on each smaller ball $B'\Subset B$ the mollifications $v_\varepsilon $ and $\partial _ av_\varepsilon =(\partial _ av)_\varepsilon $ converge uniformly to the continuous representatives $\tilde v$ of $v$ and $\tilde w_ a$ of $\partial _ av$. Passing to the limit in

shows that $\tilde v$ has the continuous partial derivatives $\partial _ a\tilde v=\tilde w_ a$. Hence $\tilde v$ is continuously differentiable, and by the first part its first derivatives are Hölder continuous with exponent $\alpha $ on each $\overline{B'}$. Every compact subset of $U$ is covered by finitely many such balls $B'$; pairs of points closer than a Lebesgue number of this cover lie in a common ball, and for the remaining pairs the supremum bound suffices. Hence $\tilde v\in C^{1,\alpha }_{\mathrm{loc}}(U)$. $\square$

Proof. Identify the space of complex $n\times n$ matrices with $\mathbb R^{2n^2}$ through the real and imaginary parts of the entries, and let $\gamma $ be the measure on it with density $\mathrm e^{-\pi \operatorname {tr}(X^*X)}$ with respect to Lebesgue measure. Since $\operatorname {tr}(X^*X)=\sum _{j,k}|X_{jk}|^2$, this density is the Gaussian $\mathrm e^{-\pi |x|^2}$ on $\mathbb R^{2n^2}$, so $\gamma $ is a probability measure by lemma 2.9.2 (with $\xi =0$ and $t=1$). The singular matrices form a $\gamma $-null set: they are the zero set of the real polynomial $|\det X|^2$, which is not identically zero since $\det 1=1$, and the zero set of such a polynomial is Lebesgue null.

To prove the last fact, let $P$ be a real polynomial in $m\geq 1$ variables that is not identically zero; we argue by induction on $m$. If $m=1$, the zero set is finite. If $m\geq 2$, write $P=\sum _ kc_ k(x')x_ m^ k$ as a polynomial in the last variable, with coefficients $c_ k$ that are polynomials in $x'=(x_1,\ldots ,x_{m-1})$. Some $c_ k$ is not identically zero, so by induction the common zero set $Z'$ of the coefficients is null in $\mathbb R^{m-1}$. For $x'\notin Z'$, the polynomial $P(x',\cdot )$ is not identically zero and has finitely many zeros. By Tonelli's theorem applied to the indicator function of the closed set $\{ P=0\} $, this set is null.

For invertible $X$ define

where $(X^*X)^{1/2}$ is the positive square root of the positive definite matrix $X^*X$. By the spectral theorem for Hermitian matrices, this square root exists and is unique. It depends continuously on the matrix. Indeed, let $A_ k\to A$ be positive definite matrices. The square roots $A_ k^{1/2}$ are bounded, since $\operatorname {tr}(A_ k^{1/2}A_ k^{1/2})=\operatorname {tr}A_ k$. Every convergent subsequence of them has a positive semidefinite limit $S$ with $S^2=A$; then $S$ is invertible, hence positive definite, and $S=A^{1/2}$ by uniqueness. A bounded sequence of matrices all of whose convergent subsequences have the same limit converges to it, so $A_ k^{1/2}\to A^{1/2}$. Since inversion is continuous on invertible matrices, $Q$ is continuous on the open set of invertible matrices, and $Q(X)^*Q(X)=(X^*X)^{-1/2}X^*X(X^*X)^{-1/2}=1$. Let $\eta $ be the image under $Q$ of the restriction of $\gamma $ to the invertible matrices; it is a Borel probability measure on $\mathrm U(n)$.

Fix $U,V\in \mathrm U(n)$. The map $X\mapsto UXV$ is real linear and preserves $\operatorname {tr}(X^*X)$, so it is orthogonal on $\mathbb R^{2n^2}$ and preserves $\gamma $. By uniqueness of positive square roots, $(V^*X^*XV)^{1/2}=V^*(X^*X)^{1/2}V$, and therefore $Q(UXV)=UQ(X)V$. Consequently the image of $\eta $ under $g\mapsto UgV$ is the image of $\gamma $ under $X\mapsto Q(UXV)$, which is $\eta $ because $X\mapsto UXV$ preserves $\gamma $. This proves the invariance of $\eta $ on both sides.

Now let $Y$ be as in the statement and fix $y_0\in Y$. By transitivity, $Y$ is the image of the compact space $\mathrm U(n)$ under the continuous map $g\mapsto gy_0$, so $Y$ is compact, and hence separable. Let $\sigma $ be the image of $\eta $ under $g\mapsto gy_0$. For $h\in \mathrm U(n)$, the image of $\sigma $ under $h$ is the image of $\eta $ under $g\mapsto hgy_0$, which is $\sigma $ by left invariance; thus $\sigma $ is invariant. Let $f$ be continuous on $Y$. Every $y\in Y$ has the form $hy_0$, by transitivity, and right invariance gives

Let $\tau $ be any invariant Borel probability measure on $Y$. We integrate the last identity against $\tau $ and apply Fubini's theorem to the bounded continuous function $(g,y)\mapsto f(gy)$ on $\mathrm U(n)\times Y$, which is product-measurable by lemma 2.2.8 because $\mathrm U(n)$ and $Y$ are separable metric spaces. We obtain

Thus $\sigma $ and $\tau $ agree on continuous functions. For a nonempty closed set $F\subset Y$, the continuous functions $\max \{ 0,1-j\operatorname {dist}(y,F)\} $ decrease to $\mathbf1_ F$, so dominated convergence gives $\sigma (F)=\tau (F)$. The closed sets are stable under finite intersections and generate the Borel $\sigma $-algebra, so lemma 2.2.6(2) gives $\sigma =\tau $. In particular $\sigma $ does not depend on $y_0$.

Finally, $\mathrm U(n)$ acts continuously on the unit sphere of $\mathbb C^ n$, and transitively: every unit vector is the first column of a unitary matrix, obtained by completing it to an orthonormal basis with the Gram–Schmidt process. For the Grassmannians, these properties are verified below, where they are introduced. $\square$

Proof. If $n=1$, then $\mathbb P^0(\mathbb C)$ is a single point, and both sides equal $1$; so let $n\geq 2$. Let $\gamma $ be the probability measure on $\mathbb C^ n=\mathbb R^{2n}$ with density $\mathrm e^{-\pi |z|^2}$. Unitary maps are orthogonal on $\mathbb R^{2n}$, so they preserve Lebesgue measure and $|z|$; hence $\gamma $ is unitarily invariant. Moreover $\gamma (\{ 0\} )=0$. The map $z\mapsto \mathbb Cz$ from $\mathbb C^ n\setminus \{ 0\} $ to $\mathbb P^{n-1}(\mathbb C)$, that is, $z\mapsto zz^*/|z|^2$, is continuous and commutes with the action of $\mathrm U(n)$. Hence the image of $\gamma $ under this map is a unitarily invariant Borel probability measure, and by the uniqueness in proposition 2.9.18 it is $\sigma $.

The metric $d$ is unitarily invariant, because unitary matrices preserve $\langle \cdot ,\cdot \rangle $. Since $\sigma $ is invariant and the unitary group acts transitively, we may therefore take $\xi _0=\mathbb Ce_1$, the first coordinate line. Write $z=(z_1,z')$ with $z'\in \mathbb C^{n-1}$. For the line $\xi =\mathbb Cz$, the condition $d(\xi ,\xi _0)<a$ means $|z_1|>\cos a\, |z|$, because $\arccos $ is strictly decreasing; that is, $|z_1|^2\sin ^2a>\cos ^2a\, |z'|^2$, or $|z_1|^2>\cot ^2a\, |z'|^2$, where $\cot (\pi /2)=0$. Thus the preimage of the ball under $z\mapsto \mathbb Cz$ is the set of $z\neq 0$ with $|z_1|^2>\cot ^2a\, |z'|^2$. We compute its $\gamma $-measure by Fubini's theorem, integrating first in $z_1$. For $c\geq 0$, polar coordinates in $\mathbb C$ give

and we apply this with $c=\cot ^2a\, |z'|^2$. Using lemma 2.9.2 in $\mathbb C^{n-1}=\mathbb R^{2n-2}$ with $\xi =0$ and $t=1+\cot ^2a$, we obtain

The last assertion follows from $\sin a\leq a$. $\square$

Proof. Write $|\cdot |$ for Lebesgue measure on $[0,2\pi ]$, and put $\ell (t)=|F_ t|$ and $A_ t(b)=|F_ t\cap [0,b]|$ for $b\in [0,2\pi ]$. We first show that, for fixed $b$, the function $t\mapsto A_ t(b)$ is Borel; with $b=2\pi $, so is $\ell $. Since $T$ and $[0,2\pi ]$ are separable metric spaces, the Borel sets of $T\times [0,2\pi ]$ form the $\sigma $-algebra generated by the products of Borel sets (lemma 2.2.8). Hence the sections $G_ t$ of a Borel set $G$ are Borel, because the sets all of whose sections are Borel form a $\sigma $-algebra containing these products. The class of Borel sets $G\subset T\times [0,2\pi ]$ for which $t\mapsto |G_ t\cap [0,b]|$ is Borel contains the products $B_1\times B_2$ of Borel sets, for which this function is $\mathbf1_{B_1}(t)|B_2\cap [0,b]|$. It is closed under complements, since the complement of $G$ gives $b-|G_ t\cap [0,b]|$, and under countable disjoint unions, which give sums. Since the products of Borel sets are closed under finite intersections, lemma 2.2.6(1) shows that the class contains every Borel set.

For fixed $t$, the function $A_ t$ is nondecreasing and $1$-Lipschitz, with $A_ t(0)=0$ and $A_ t(2\pi )=\ell (t)>0$. Define

Because $A_ t$ is continuous and nondecreasing and $0<\ell (t)/2<\ell (t)$, the set of all $b\in [0,2\pi ]$ with $A_ t(b)\geq \ell (t)/2$ is an interval $[b_ t,2\pi ]$ with $0<b_ t<2\pi $ and $A_ t(b_ t)=\ell (t)/2$. The rational numbers are dense in this interval, so $a(t)=b_ t$ and $A_ t(a(t))=\ell (t)/2$. The map $a$ is Borel, since for every $c\in (0,2\pi ]$

Finally, suppose that $a(t)\notin F_ t$. Since $F_ t$ is closed and $a(t)>0$, some interval $[a(t)-\varepsilon ,a(t)]$ with $\varepsilon >0$ misses $F_ t$, and $A_ t$ is constant on it. A rational number $b$ in this interval with $b<a(t)$ then satisfies $A_ t(b)=\ell (t)/2$, contradicting the definition of $a(t)$. Hence $a(t)\in F_ t$. $\square$

Proof. We obtain $\mu $ by choosing, in a Borel way, a point of $E$ on each circle of radius $r\in [R/2,R]$ centered at $x$ in a complex line through $x$, and then we estimate its energy. For $\xi \in \mathbb P^{n-1}(\mathbb C)$ we first choose a unit vector $e_0(\xi )\in \xi $ in a Borel way: if $j$ is the first index with $(P_\xi )_{jj}>0$, put $e_0(\xi )=P_\xi e_ j/\sqrt{(P_\xi )_{jj}}$. Since $|P_\xi e_ j|^2=\langle P_\xi e_ j,e_ j\rangle =(P_\xi )_{jj}$, this is a unit vector in $\xi $. The set of $\xi $ with a given first index $j$ is Borel, and $e_0$ is continuous on it.

Let $r\in [R/2,R]$ and $\xi \in \mathbb P^{n-1}(\mathbb C)$. The circle $\theta \mapsto x+r\mathrm e^{\mathrm i\theta }e_0(\xi )$ lies in the closed disc $\{ x+\zeta e_0(\xi ):|\zeta |\leq r\} $, which is connected, contains $x$ and is contained in $D$; hence it lies in the connected component containing $x$ of the section of $D$ by the line $x+\xi $. On this component $u$ is subharmonic; in particular it is upper semicontinuous and bounded above on the circle, and by theorem 3.4.5(2) and the arc-length formula for circle means at the beginning of this section, it satisfies the submean inequality

Let

This set is closed by upper semicontinuity, and it has positive length. If $A=-\infty $, it is the whole interval $[0,2\pi ]$. If $A$ is finite and $F_{r,\xi }$ were a null set, then $A-u$ would be positive almost everywhere on the circle, so the circle mean of $u$ would be strictly less than $A$, contradicting the submean inequality.

The set of $(r,\xi ,\theta )$ with $\theta \in F_{r,\xi }$ is Borel, because $u$ is Borel and the map $(r,\xi ,\theta )\mapsto x+r\mathrm e^{\mathrm i\theta }e_0(\xi )$ into $\overline{B(x,R)}$ is Borel, being continuous on each of the finitely many Borel sets on which $e_0$ is continuous. Since $[R/2,R]\times \mathbb P^{n-1}(\mathbb C)$ is a compact, hence separable, metric space, lemma 2.9.21 gives a Borel map $(r,\xi )\mapsto \alpha (r,\xi )$ with $\alpha (r,\xi )\in F_{r,\xi }$. Put $e(r,\xi )=\mathrm e^{\mathrm i\alpha (r,\xi )}e_0(\xi )$ and $\Psi (r,\xi )=x+re(r,\xi )$. Then $\Psi $ is Borel, hence product-measurable (lemma 2.2.8), and it takes its values in $E$. Let $\mu $ be the image under $\Psi $ of the product of normalized Lebesgue measure on $[R/2,R]$ and $\sigma $. It is a Borel probability measure carried by $E$.

It remains to estimate the energy; the estimate does not depend on the choice of the phases $\alpha (r,\xi )$. For $r,r'\in [R/2,R]$ and unit vectors $e,e'$ spanning $\xi ,\xi '$,

where the identity follows by expanding both sides with $|e|=|e'|=1$ and $|e-e'|^2=2-2\operatorname{Re}\langle e,e'\rangle $, and the inequality holds since $rr'\geq R^2/4$ and $\operatorname{Re}\langle e,e'\rangle \leq |\langle e,e'\rangle |$. With $a=d(\xi ,\xi ')\in [0,\pi /2]$, we have $2-2|\langle e,e'\rangle |=4\sin ^2(a/2)\geq 4a^2/\pi ^2$, because $\sin y\geq 2y/\pi $ for $0\leq y\leq \pi /2$ by concavity of the sine on this interval. Hence $|re-r'e'|^2\geq (r-r')^2+R^2a^2/\pi ^2$.

Write $r=Rt$ with $t\in [1/2,1]$. Equip $[1/2,1]\times \mathbb P^{n-1}(\mathbb C)$ with the metric $\rho $ defined by $\rho ((t,\xi ),(t',\xi '))^2=(t-t')^2+d(\xi ,\xi ')^2$, and with the product probability $m$ of normalized Lebesgue measure and $\sigma $, which is defined on all Borel sets of this separable metric space by lemma 2.2.8. Then $\mu $ is the image of $m$ under $(t,\xi )\mapsto \Psi (Rt,\xi )$, and the lower bound for $|re-r'e'|^2$, together with $\pi >1$, gives

A $\rho $-ball of radius $s\leq 1$ is contained in the product of an interval of length $2s$, whose normalized measure is at most $4s$, and a ball of radius $s$ for the metric $d$. Since $s\leq 1<\pi /2$, lemma 2.9.19 shows that its $m$-measure is at most $4s^{2n-1}$. In particular $m$ has no atoms, so the diagonal, a closed and hence product-measurable set (lemma 2.2.8), is null for $m\times m$ by Tonelli's theorem. Fix $z\in [1/2,1]\times \mathbb P^{n-1}(\mathbb C)$. Splitting the integral into the set where $\rho (z,z')\geq 1$ and the shells where $2^{-k-1}\leq \rho (z,z')<2^{-k}$, $k\geq 0$, we obtain

Since $\mu $ is the image of $m$, the product $\mu \times \mu $ is the image of $m\times m$ under the product map, because both measures agree on rectangles (proposition 2.2.7). Therefore the lower bound for $|\Psi (Rt,\xi )-\Psi (Rt',\xi ')|$, valid off the $m\times m$-null diagonal, and an integration in $z$ give

This is the assertion, with $C_ n=\pi ^{2n-2}(1+2^{2n+1})$. $\square$

Proof. We write the inner integral in polar coordinates on $H$ and show that the normalized surface measures of the unit spheres of the hyperplanes $H$ average to the normalized surface measure of $S^{2n-1}$. For $H\in \mathcal H$, let $S_ H$ be the unit sphere of $H$, and let $\tau _ H$ be the image under $T_ H$ of the normalized surface measure $\mathrm dS/s_{2n-3}$ of the unit sphere $S^{2n-3}$ of $\mathbb C^{n-1}=\mathbb R^{2n-2}$. We regard $\tau _ H$ as a Borel probability measure on $S^{2n-1}$ carried by $S_ H$; it does not depend on the frame, since surface measure is invariant under orthogonal transformations (subsection 3.1.1). Polar coordinates (3.1.2) on all of $\mathbb C^{n-1}$, transported to $H$ by $T_ H$, give

Measurability. Let $h$ be a nonnegative Borel function on $(0,\infty )\times S^{2n-1}$. By the definition of $\tau _ H$,

The map $(H,r,\omega )\mapsto (r,T_ H\omega )$ is Borel by the frame construction of subsection 2.9.17, since its two components are Borel. Hence $(H,r,\omega )\mapsto h(r,T_ H\omega )$ is a nonnegative Borel function on the product of the separable metric spaces $\mathcal H\times (0,\infty )$ and $S^{2n-3}$, and by lemma 2.2.8 and Tonelli's theorem the right side of (2.9.25) is a Borel function of $(H,r)$. In particular, taking $h(r,\theta )=g(\theta )$, we see that $H\mapsto \int g\, \mathrm d\tau _ H$ is Borel for every nonnegative Borel function $g$ on $S^{2n-1}$; for instance, $H\mapsto \tau _ H(B)$ is Borel for every Borel set $B\subset S^{2n-1}$.

Averaging. For a Borel set $B\subset S^{2n-1}$ put $\tau (B)=\int \tau _ H(B)\, \mathrm d\nu (H)$. By monotone convergence, $\tau $ is countably additive, so it is a Borel probability measure on $S^{2n-1}$. For a unitary matrix $U$, the map $UT_ H$ is a complex linear isometry from $\mathbb C^{n-1}$ onto $UH$, so $U$ maps $\tau _ H$ to $\tau _{UH}$, that is, $\tau _ H(U^{-1}B)=\tau _{UH}(B)$. Since $\nu $ is invariant, it follows that $\tau (U^{-1}B)=\int \tau _{UH}(B)\, \mathrm d\nu (H)=\tau (B)$, so $\tau $ is unitarily invariant. The normalized surface measure $\mathrm dS/s_{2n-1}$ of $S^{2n-1}$ is also unitarily invariant, because unitary matrices are orthogonal transformations of $\mathbb R^{2n}$. By the uniqueness in proposition 2.9.18, $\tau =\mathrm dS/s_{2n-1}$. By the definition of $\tau $ for indicator functions, by linearity for simple functions, and by monotone convergence in general, $\int _{\mathcal H}\int g\, \mathrm d\tau _ H\, \mathrm d\nu (H)=\int g\, \mathrm d\tau $ for every nonnegative Borel function $g$ on $S^{2n-1}$.

Conclusion. Integrate the polar-coordinate identity over $H$ with respect to $\nu $. The function $(r,\theta )\mapsto f(r\theta )$ is Borel on $(0,\infty )\times S^{2n-1}$, so by the measurability step $(H,r)\mapsto r^{2n-3}\int f(r\theta )\, \mathrm d\tau _ H(\theta )$ is a nonnegative Borel function on $\mathcal H\times (0,\infty )$, which is product-measurable by lemma 2.2.8. Tonelli's theorem therefore allows us to integrate first over $H$, and the averaging identity with $g(\theta )=f(r\theta )$ shows that the left side of the lemma equals

In polar coordinates $\xi =r\theta $ on all of $\mathbb C^ n=\mathbb R^{2n}$, (3.1.2) gives $|\xi |^{-2}\, \mathrm d\xi =r^{2n-3}\, \mathrm d r\, \mathrm dS(\theta )$, so this is the right side. All integrals may be infinite. $\square$

Proof. On $H$ we use the Fourier transform defined by the Euclidean inner product $\operatorname{Re}\langle \cdot ,\cdot \rangle $ and the measure $m_ H$; in an orthonormal frame it is the transform of subsection 2.9.1 on $\mathbb R^{2n-2}$. For $y\in \mathbb C^ n$ and $\xi \in H$ we have $\operatorname{Re}\langle P_ Hy,\xi \rangle =\operatorname{Re}\langle y,\xi \rangle $, because $y-P_ Hy$ is orthogonal to $H$. Hence the Fourier transform of $(P_ H)_*\mu $ at $\xi \in H$ is $\int \mathrm e^{-2\pi \mathrm i\operatorname{Re}\langle P_ Hy,\xi \rangle }\, \mathrm d\mu (y)=\widehat\mu (\xi )$; that is, it is the restriction of $\widehat\mu $ to $H$. Apply lemma 2.9.24 to the continuous function $|\widehat\mu |^2$, and then theorem 2.9.6 with $d=2n$ and $s=2n-2$, for which $|\xi |^{s-d}=|\xi |^{-2}$. We obtain

The inner integral is a Borel function of $H$, by the frame construction in subsection 2.9.17; hence it is finite for $\nu $-almost every $H$. For each such $H$, we apply proposition 2.9.5 on $\mathbb C^{n-1}=\mathbb R^{2n-2}$ to the image of $\mu $ under $T_ H^{-1}P_ H$. Since $T_ H$ preserves $\operatorname{Re}\langle \cdot ,\cdot \rangle $, the Fourier transform of this image is $w\mapsto \widehat\mu (T_ Hw)$, whose squared $L^2$ norm is $\int _ H|\widehat\mu |^2\, \mathrm d m_ H<\infty $. Hence the image has a density in $L^1\cap L^2$, and composing it with $T_ H^{-1}$ we obtain a density $g_ H$ of $(P_ H)_*\mu $ with respect to $m_ H$, with $\| g_ H\| _2^2=\int _ H|\widehat\mu |^2\, \mathrm d m_ H$. In particular, $H\mapsto \| g_ H\| _2^2$ is Borel, being equal to this Fourier integral, so no measurable choice of the densities themselves is needed. Integrating over $H$ proves the identity.

To compute $\beta _ n$, recall from (2.9.8) that $s_{k-1}=2\pi ^{k/2}/\Gamma (k/2)$ for every $k\geq 1$. Hence $a_ n=\Gamma (n)/(\pi \Gamma (n-1))$, while $b_{2n,2n-2}=\pi ^{n-2}/\Gamma (n-1)$, and therefore $\beta _ n=\Gamma (n)/\pi ^{n-1}$. $\square$

Proof. We parametrize the affine complex lines that can meet $K$ by a compact set carrying a finite measure $\lambda $, and bound from below, independently of $j$, the $\lambda $-measure of a set of lines meeting $E_ j$. Choose $M>0$ with $K\subset B(0,M)$. Every affine complex line can be written as $c+H^\perp $ with $H\in \mathcal H$ and $c\in H$, and $c=P_ Hy$ for every point $y$ of the line. Indeed, if the line is $a+\mathbb Cb$, let $H$ be the hyperplane orthogonal to $b$, so that $H^\perp =\mathbb Cb$, and put $c=P_ Ha$; then $P_ H(a+\zeta b)=c$ for every $\zeta \in \mathbb C$. If the line meets $K$ at a point $y$, then $|c|=|P_ Hy|\leq |y|<M$. Hence the lines that can meet $K$ are parametrized by the set

which is compact, being closed in $\mathcal H\times \overline{B(0,M)}$ because $c\in H$ means $P_ Hc=c$. We equip $\mathcal L_ M$ with the finite measure $\lambda $ defined by

for nonnegative Borel functions $h$. In the frames of subsection 2.9.17, $\lambda $ is the image of the product of $\nu $ and Lebesgue measure on the closed ball of radius $M$ in $\mathbb C^{n-1}$. Thus it is a Borel measure, and its total mass is the volume of a ball of radius $M$ in $\mathbb R^{2n-2}$.

By inner regularity (proposition 2.2.2), choose compact sets $F_ j\subset E_ j$ with $\mu _ j(F_ j)\geq 1/2$. The set

parametrizes the lines meeting $F_ j$. It is compact, as the image of $\mathcal H\times F_ j$ under the continuous map $(H,y)\mapsto (H,P_ Hy)$. Its section over $H$ is $P_ HF_ j$, so $\lambda (J_ j)=\int m_ H(P_ HF_ j)\, \mathrm d\nu (H)$. Let $g_{j,H}$ be the densities given by theorem 2.9.26 for $\mu _ j$. Since $F_ j\subset P_ H^{-1}(P_ HF_ j)$, the Cauchy–Schwarz inequality gives, for $\nu $-almost every $H$,

Integrate this inequality over $H$ with respect to the probability measure $\nu $, and apply the Cauchy–Schwarz inequality once more. Squaring the result and using theorem 2.9.26, we obtain

In particular $C>0$, and $\lambda (J_ j)\geq \delta =1/(4\beta _ n C)$ for every $j$. The sets $\bigcup _{j\geq m}J_ j$ decrease in $m$ and have measure at least $\delta $. Since $\lambda $ is finite, continuity from above gives

so this set is nonempty. If $(H,c)$ belongs to it, then $c\in P_ HF_ j$ for infinitely many $j$. For each such $j$ there is $y\in F_ j$ with $c=P_ Hy$, so $y\in c+H^\perp $. Hence the line $c+H^\perp $ meets $F_ j$, and therefore $E_ j$, for infinitely many $j$. $\square$

Proof. If $n=1$, the only affine complex line is $\mathbb C$ itself. Thus $u$ is subharmonic on every component of $D$, hence upper semicontinuous and locally bounded above.

Let $n\geq 2$, and suppose that $u$ is not bounded above on any neighborhood of some point $p\in D$. Then there are points $x_ j\to p$ with $A_ j=u(x_ j)\to \infty $. Choose $R>0$ with $\overline{B(p,2R)}\subset D$; discarding finitely many terms, we may assume that $|x_ j-p|<R$ for all $j$. Then $\overline{B(x_ j,R)}\subset B(p,2R)\subset D$, and theorem 2.9.22 gives Borel probability measures $\mu _ j$ carried by the sets

with $I_{2n-2}(\mu _ j)\leq C_ nR^{-(2n-2)}$ for every $j$. All the sets $E_ j$ lie in the bounded set $B(p,2R)$. By theorem 2.9.27, some affine complex line $L$ meets $E_ j$ for all $j$ in an infinite set $J$. Choose $y_ j\in L\cap E_ j$ for $j\in J$. Passing to a subsequence, we may assume that $y_ j\to y\in L\cap \overline{B(p,2R)}\subset D$. A small closed disc in $L$ centered at $y$ lies in $D$, so for large $j$ the points $y_ j$ lie in the same connected component of $D\cap L$ as $y$. On this component $u$ is subharmonic, hence upper semicontinuous, and therefore $\varlimsup _ ju(y_ j)\leq u(y)<\infty $. This contradicts $u(y_ j)\geq A_ j\to \infty $. $\square$